Complex Integer Solutions of x, y nu, v(n = 1, 5)
Python之复数、分数、大型数组数学运算（complex、cmath、numpy、fra。。。
Python之复数、分数、⼤型数组数学运算（complex、cmath、numpy、fra。
⼀、复数的数学运算复数可以⽤使⽤函数 complex(real, imag) 或者是带有后缀j的浮点数来指定a=complex(2,4)print(a) # (2+4j)b=2-5j# 获取对应的实部、虚部和共轭复数print(b.real,b.imag,b.conjugate()) # 2.0 -5.0 (2+5j)# 数学运算print(a+b) # (4-1j)print(abs(a)) # 4.47213595499958复数函数⽐如正弦、余弦或平⽅根，使⽤ cmath 模块import cmathprint(cmath.sin(a))print(cmath.cos(a))print(cmath.exp(a))# (24.83130584894638-11.356612*********j)# (-11.36423470640106-24.814651485634183j)# (-4.829809383269385-5.5920560936409816j)使⽤ numpy 也可以处理复数# 构造复数数组import numpy as npc=np.array([1+2j,2+3j,3+4j])print(c) # [ 1.+2.j 2.+3.j 3.+4.j]print(c+2) # [ 3.+2.j 4.+3.j 5.+4.j]print(np.sin(c)) # [ 3.16577851 +1.95960104j 9.15449915 -4.16890696j 3.85373804-27.01681326j]补充：# Python的标准数学函数确实情况下并不能产⽣复数值,必须使⽤cmathprint(cmath.sqrt(-1)) # 1j⼆、分数运算fractions 模块可以被⽤来执⾏包含分数的数学运算from fractions import Fractiona=Fraction(1,3)b=Fraction(1,4)print(a,b,a+b) # 1/3 1/4 7/12c=a*bprint(c) # 1/12三、⼤型数组运算⾸先了解：标准列表对象和 NumPy 数组对象之间的差别标准列表对象：x = [1, 2, 3, 4]y = [5, 6, 7, 8]print(x * 2) # [1, 2, 3, 4, 1, 2, 3, 4]# print(x + 10) # 报错了TypeError: can only concatenate list (not "int") to listprint(x + y) # [1, 2, 3, 4, 5, 6, 7, 8]Numpy arraysimport numpy as npax = np.array([1, 2, 3, 4])ay = np.array([5, 6, 7, 8])print(ax * 2) # array([2, 4, 6, 8])print(ax + 10) # array([11, 12, 13, 14])print(ax + ay) # array([ 6, 8, 10, 12])print(ax * ay) # array([ 5, 12, 21, 32])计算多项式的值,即每个列表元素进⾏多项式运算def f(x):return 3*x**2 - 2*x + 7print(f(ax)) # [ 8 15 28 47]多维数组，索引问题，列表基层嵌套，就是⼏维数组，索引：逗号分不同维度的# eg (1:3,2:3) ⼆维数组中取第⼀层索引1到3，再取索引1到3列表中索引的2:4a = np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]) # ⼆维数组print(a[1:3,1:3]) # [[ 6 7] [10 11]]print(a[:,1]) # [ 2 6 10]print(a+2) # [[ 3 4 5 6] [ 7 8 9 10] [11 12 13 14]]补充：numpy 优势：numpy中的函数要⽐循环数组并使⽤ math 模块中的函数执⾏计算要快的多。
imc2000-day2-solutions(IMC国际数学竞赛题目2000第2天)
y = a0x4 + a1x3 + a2x2 + a3x + a4 x2 + a5x,
where the equation y = a5x determines the straight line A1A3. The abscissas of the points A1 and A3 are −a and a, a > 0, respectively. Since −a and a are points of tangency, the numbers −a and a must be double roots of the polynomial a0x4 + a1x3 + a2x2 + a3x + a4. It follows that the polynomial is of the form
µ(p, c) − µ(p c) = 1 or µ(p − 1, c) − µ(p c) = 1,
(3)
respectively. Putting (1), (2) and (3) together we obtain
S0 + S1 =
µ(p, c) − µ(p , c) +
µ(p − 1, c) − µ(p , c) =
Now observe that any dissection of a cube into n smaller cubes may be refined to give a dissection into n + (ad − 1) cubes, for any a ≥ 1. This refinement is achieved by picking an arbitrary cube in the dissection, and cutting it into ad smaller cubes. To prove the required result, then, it suffices to exhibit two relatively prime integers of form ad − 1. In the 2-dimensional case, a1 = 2 and a2 = 3 give the coprime numbers 22 − 1 = 3 and 32 − 1 = 8. In the general case, two such integers are 2d − 1 and (2d − 1)d − 1, as is easy to check.
关于Diophantine方程4x2n－py2＝1
关于Diophantine方程4x2n－py2＝1贺艳峰;柴璇【摘要】Let n be a positive integer with n>1 ,and let p be an odd prime with p>3 .By using the elementary and the properties of the generalized Lebesgue‐Nagell equations and the general‐ized Fermat equations ,the integer solutions of the equation 4 x2n -p y2 =1 is studied .It is prove that this equation has no positive integer solution (x ,y) for 2︳ n .%设 n ＞1是正整数，p 是大于3的奇素数。
本文运用初等数论的方法，结合广义Lebesgue‐Nagell方程和广义Fermat方程的性质，研究了丢番图方程4 x2n － py2＝1的整数解，并证明了对于任意奇数 n ，此方程没有正整数解（x ，y）。
【期刊名称】《纺织高校基础科学学报》【年(卷),期】2015(000)001【总页数】3页(P45-47)【关键词】高次Diophantine方程;广义 Lebesgue-Nagell方程;广义 Fermat方程【作者】贺艳峰;柴璇【作者单位】延安大学数学与计算机科学学院，陕西延安716000;延安大学数学与计算机科学学院，陕西延安716000【正文语种】中文【中图分类】O156.7设Z,N,P分别是全体整数, 正整数以及大于3的奇素数的集合. 本文讨论方程的求解问题. 这是一类基本而又重要的高次Diophantine方程, 对此类方程的研究参见文献[1-6].文献[7]证明了: 方程(1)仅有解(p,x,y,n)=(7,2,3,2) 适合n=2.根据这一结果可以确定该方程的所有适合 2|n的解 (p,x,y,n).文献[8]再次证明了上述结果. 本文对于 2⫮n时的情况证明了以下结果:定理1 方程(1)没有适合 2⫮n的解 (p,x,y,n).至此, 方程(1)的求解问题全部解决.为了证明定理1,首先引入如下两个引理:引理1 方程无解 (q,X,Y).证明参阅文献[9].引理2 对于奇素数q, 方程没有适合 1<r<q的解 (X,Y,Z,r).证明参阅文献[10].设 (p,x,y,n) 是方程 (1) 的一组适合 2⫮n的解. 因为n>1, 所以n必有奇素因子q. 设将式(4)代入(1), 即得由于2⫮py2, 而且gcd(2zq+1, 2zq-1)=1, 故从式(5)可得从式(6)可得以及将式(8) 代入式(7), 可得当λ=1 时, 从式(9)可知因为2⫮y, 故从式(6)可知2⫮a, 所以8|a2-1. 因此, 从式(10) 可知又因gcd(a+1,a-1)=2, 故从式(10)和(11)可得从式(12)可得由于q是奇素数, 故从式(13) 可得从式(14)可知, 此时方程(3)有解(X,Y,Z,r)=(f,-g,-λ′,q-1). 然而, 因为q≥3 且 1<q-1<q, 所以根据引理2可知这是不可能的.当λ=-1时, 从式(12)可得从式(15)可知方程(2)有解(X,Y)=(a,z).然而, 因为q是奇素数, 所以根据引理1可知a=z=1, 并且从式(7)可得p=3.由此可知, 这是不可能的.综上所述,方程(1)没有适合2⫮n的解 (p,x,y,n). 定理1证毕.【相关文献】[1] MORDELLLJ.Diophantineequations[M].London:AcademicPress,1969：272-276.[2] 李江华.广义Fermat商中的平方数和立方数[J].纯粹数学与应用数学,2012,28(6):774-778. LIJH.ThesquaresandcubesingeneralizedFermatquotients[J].PureApplMath,2012,28(6):774-778.[3] 刘妙华.广义Lebesgue-Nagell方程x2-4p2r=y3[J].西安工程大学学报,2013,27(6):821-823. LIUMH.ThegeneralizedLebesgue-Nagellequationx2-4p2r=y3[J].JXi′anPolyUniv,2013,27(6):821-823.[4] 王枭涵.关于Diophantine方程p2m-Dx2=1[J].纺织高校基础科学学报,2014,27(1):45-47. WANGXH.OntheDiophantineequationp2m-Dx2=1[J].BasicSciJTextUniv,2014,27(1):45-47.[5] 王建华,王枭涵.广义Brocard-Ramanujan方程x2-D=y!解的上界[J].纺织高校基础科学学报,2014,27(3):290-292.WANGJH,WANGXH.TheupperboundforsolutionsofthegeneralizedBrocard-Ramanujanequationx2-D=y![J].BasicSciJTextUniv,2014,27(3): 290-292.[6] 杨海，武静，任荣珍.关于Diophantine方程 x3-53=3py2[J].纺织高校基础科学学报,2014,27(4):418-420.YANGH,WUJ,RENRZ.TheDiophantineequationx3-53=3py2[J].BasicSciJTextUnvi,2014,27(4):418-420.[7] 曹珍富.丢番图方程引论[M].哈尔滨: 哈尔滨工业大学出版社,1989:271-272.CAOZF.IntrodutiontoDiophantineequations[M].Harbin:HarbinInstTechPress,1989:271-272.[8] 管训贵.关于不定方程4x2n-py2=1[J].湖北民族学院学报:自然科学版,2010,28(3): 341-343. GUANGX.OntheDiophantineequation4x2n-py2=1[J].JHubeiNationColl:NatSci,2010,28(3):341-343.[9]STÖMERC.Léquationmarctan(1/x)+narctan(1/y)=kπ/4[J].B ullSocMathFrance,1899,27(2):160 -170.[10] RIBETK.Ontheequationap+2abp+cp=0[J].ActaArith,1997,79(1):7-16.。
第九届哈佛大学-麻省理工数学竞赛代数题及解答
1±
1+44 2
=
1±3 2
5.
The other two solutions follow.
3. The train schedule in Hummut is hopelessly unreliable. Train A will enter Intersection X from the west at a random time between 9:00 am and 2:30 pm; each moment in that interval is equally likely. Train B will enter the same intersection from the north at a random time between 9:30 am and 12:30 pm, independent of Train A; again, each moment in the interval is equally likely. If each train takes 45 minutes to clear the intersection, what is the probability of a collision today?
is
always
1
1 2
hours
long,
so
the
probability
of
a
collision
is
1
1 2
÷
5
1 2
=
3 11
.
From 9:30 to 12:30 is 3 hours.
Now case (i) occurs with probability
关于丢番图方程1＋2x11y＋5z11u＝2v5w，xvw＞0，y＋u＞0
关于丢番图方程1＋2x11y＋5z11u＝2v5w，xvw＞0，y＋u＞0陈小燕【期刊名称】《西华师范大学学报（自然科学版）》【年(卷),期】2016(037)002【摘要】This paper discusses a special case of the Diophantine equation 1 +X +Y =Z.With computer assistance, all the nonnegative integer solutions to the exponential Diophantine equation 1 +2 x 1 1 y +5 z 1 1 u =2 v5w ,xvw >0,y +u>0 are determined by elementary method.%讨论了丢番图方程1＋X ＋Y ＝Z 的一个特殊情形。
借助计算机，用初等方法给出了指数丢番图方程1＋2 x 11 y ＋5 z 11 u ＝2 v5 w ，xvw ＞0，y ＋u ＞0的全部非负整数解。
【总页数】4页(P200-202,232)【作者】陈小燕【作者单位】琼台师范高等专科学校数理系，海南海口 571100【正文语种】中文【中图分类】O156.7【相关文献】1.关于丢番图方程p^x-q^y^z=α和丢番图方程p1xp2^y-q1zq2^w=α [J], 曹玉书2.丢番图方程x2+(2n)2=y9(1≤n≤7)的整数解 [J], 陈一维;柴向阳3.三次丢番图方程x3±33=pqy2的整数解 [J], 李恒;杨海;罗永亮4.关于丢番图方程x3+1=413y2 [J], 华程5.关于丢番图方程x3-1=1261y2 [J], 吴小英因版权原因，仅展示原文概要，查看原文内容请购买。
专业英语中常用符号和数学表达式
那你就得了解各种数学名称了，比如“y分之x”是“x over y”，“x乘以y”是“x times y”，“除以”是“divided by”，加plus, 减minus；“x的y次方”是“x to the power of y”，“根”是“root”，几次跟就是几的英语序数形式，6次根=6th root。
1.(A的三次方+B的立方根-CXD的五次方)/E=F 读做:A cube plus the cube root ofB minusC multiplied by the 5th power of D, all divided by E equals F.2.(6+3+7/12-4.35X5)/(9*1/3) 读做: Six plus three and seven twelfths minus four decimal three five multiplied by fiv e, all divided by nine and a third.3. 基本术语：数学 mathematics 数字 number 双数 even number 单数 odd number 计算 calculate口算calculate mentally 笔算calculate using pen-and-paper 竖式vertical form 一位数1-digit number两位数2-digit number 文字题word problem 应用题story problem4、常用符号： ＋ plus、 － minus、 ＝ equal(s)、＞ is greater/more than、＜ is less than ( ) brackets5、加减法及各部分名称：加法 addition 加数 addend 和 sum 减法 subtraction被减数 minuend 减数 subtrahend 差 difference6、数位知识：数位表place value chart 个 Units/Ones 十 Tens 百 Hundreds7、图形名称：长方体 cuboid 正方体 cube 圆柱 cylinder球 sphere长方形 rectangle 正方形 square三角形 triangle 圆 circle边 side 角 angle 面 face专业英语中常用符号和数学表达式1．数的分类complex number 复数zero; naught;0 零real number 实数imaginary number 虚数relational number 有理数irrational number 无理数integer number 整数positive integer number 正整数natural number 自然数negative integer number 负整数fraction 分数decimal 小数odd number 奇数even number 偶数cardinal number 基数ordinal number 序数approximate number 近似数significant number 有效数2．整数addition 加法subtraction 减法addend 加数subtrahend 减数augend 被加数minuend 被减数plus sign 加号minus sign 减号sum 和difference（diffa）差plus; add; and; increase 加minus; decrease; subtract 减is; equal 等division 除法multiplication 乘法divide 除multiply; multiplied by; times 乘divisor 除数multiplier 乘数dividend 被除数multiplicand 被乘数quotient 商product 积remainder 余数positive 正negative 负3．小数和分数numerator 分子denominator 分母decimal point 小数点naught point four 零点四fraction stroke 分数线recurring decimal 循环小数4．百分数percent 百分比interest 利息average 平均数round off 舍入discount 折扣5．幂与指数power 幂; 乘方root-extracting 开方exponent 指数logarithms 对数X squared 某数的平方cube 三次方three cubed 三次方的four 乘四次方cube root 立方根square root 平方根to rise to the power of five 使乘五次方radical sign根号6．代数algebra 代数equation 等式; 方程式inequality 不等式unknown number 未知数absolute value绝对值simple equation 一次方程quadratic equation 二次方程cubic equation 三次方程monomial 单项式polynomial 多项式variable 变量coefficient 系数differential 微分integral 积分derivative 导数function 函数ratio 比proportion 比例sign of equality等号sign of inequality不等号interval 区间matrix 矩阵theorem 定理lemma 引理definition 定义7．常用数学表达式1/2a half; one halfl/3 a third; one third2/3 two thirds1/4 a quarter; one quarter; a fourth; one fourthl/100 a（one）hundredthl/1,000 a（one）thousandth113/324 one hundred and thirteen over three hundred and twenty-four four and two-thirdsforty-five and eighty-nine over twenty-three0.1 one tenth; point one0.01 one hundredth; point zero one0.001 one thousandth; point zero zero one; point two zero one 2050.0357 two thousand and fifty point zero three five seven0.25 zero point two fivepoint two five repetend fivezero point two five recurringzero point three seven twenty-five recurring对483579四舍五入到千位round off 483579 to nearest thousand 108 one followed by eighteen zeros-30.8 negative thirty point eight2-3i two minus three i; two minus three times i2％two per cent; two percent5‰ five per mill; five permill∞infinityx＋y＝zx plus y is z; add x to y is z; x and y is z(x＋y) bracket x plus y bracket closedx-y x minus y; subtract y from x; y from x; x subtracts yx±y x plus or minus yx×y; xy xy; multiply x by y; x multiplied by y; x by y; x times y;x÷ydivide x by y; y into xx over yx : y the ration of x to yx∝y x varies as y; x is in direct proportion to yx＝y x equals y; x is equal to y; x is yx≠y x is not equal to y; x is not yx≡y x is identical to y; x is equivalent to y; x is equivalent to y;x≈y x is approximately equal to y; x approximately equals yx＞y x is greater than y; x is more than yx>>y x is much greater than y; x is far greater than yx≥y x is greater than or equal to yx＜y x is less than yx<<y x is much less than yx≤y x is less than or equal to y0＜x＜1 zero is less than x is less than 1; x is greater than zero and less than 1 0≤x≤1 zero is less than or equal to x is less than or equal to 1x2 x square; x squared; the square of x;the second power of x; x to second powerx3 x cube; x cubed; the cube of x;the third power of x; x to the third powerxn the nth power of x; x to the nth power; x to the power nthe square root of x; x squaredthe cube root of xthe nth root of xx-n x to the (power) minus n(x+y)2x plus y all squaredx over y all squaredx i x i; x subscript i; x suffix i; x sub ilognxlog x to the base n; log of x to the base nlog10x log x to base 10; common logarithmlogex; lnx log x to the base e; log to the base e of y; natural log (of) yex; exp(x) exponential function of x, e to the power xthe summation of x sub i, where i goes from 1 to n;the sum from i equals one to n x i;the sum as i runs from one to n of the x ithe product of x sub i, where i goes from one to nthe product of all x i from i equals one to nthe product of all xi from i equals one to infinitythe absolute value of x; mod x; modulus xthe mean value of x; x barx hatx tildex* x asteriskx primex double primex double prime sub mf(x) f x; f of x; the function f of xa function f from S to Tf¢(x) f prime x; f dash x; the (1st) derivative of f with respect to xf²(x) f double–prime x; f double–dash x; the second derivative of f with respect to x f¢²(x) f triple–prime x; f triple–dash x; the third derivative of f with respect to x f4(x) four x; the fourth derivative of f with respect to xx! n factorial△finite difference or increment△x,δx the increment of xdx dee x; dee of x; differential xdel; nablanth del (nabla)the differential coefficient of y with respect to x;the first derivative of y with respect of xthe second derivative of y with respect of xthe nth derivative of y with respect of xthe partial (derivative) of y with respect to uthe second partial (derivative) of y with respect to xthe partial derivation of z with respect to x of the partial derivative of z with respect to y ∫integral of∫∫double Integral of∫…∫n-fold integral ofthe integral between limits a and b; the integral from a to bthe indefinite integral of a times x with respect to xthe integral from a to b of function of xthe double integral of f of x,ythe limit as x approaches 0the limit as x approaches 0 from abovethe limit as x approaches 0 from belowthere existsfor all∵because∴thereforex⊥y x is perpendicular to yx∥y x is parallel to yx～y the difference between x and yx∝y x varies directly as yxÞy x implies y; if x, then yxÛy x if and only if y; x is equivalent to y; x and y are equivalent { };empty setxÎA x belongs to A; x is an element (or a member) of AxÏA x does not belong to A; x is not an element (or a member) of AAÌB A is contained in B; A is a subset of BAÉB A contains B;B is a subset of AAÇB A cap B; A meet B; A intersection BAÈB A cup B; A join B; A union BA\B A minus B; the difference between A and BA×B A cross B; the Cartesian product of A and B(A与B的笛卡尔积)||A||the norm (or modulus) of Avector FAB; the length of the segment ABAT A transpose; the transpose of AA-1 A inverse; the inverse of Ax→y x maps into y; x is sent (or mapped) to yx→∞x approaches infinity∠xangle xx is perpendicular to yx is parallel to ysin sinecos cosinetg, tan tangentctg, cot cotangentsc, sec secantcsc, cosec cosecantsin-1, arcsin arc sinecos-1, arcos arc cosinesinh the hyperbolic sinecosh the hyperbolic cosine( ) round brackets; parentheses ;the signs of grouping [ ] square（angular）brackets; bracket< > angle bracket{} braces8．常用希腊字母字母读音字母读音字母读音alphabeta ,gamaxi ,psideltaepsilonzeta ,phiomegalambdamnnuetarho,sigmataupi9．其他数学名词line 线angle 角intersecting line 相交线parallel line 平行线triangle 三角形quadrilateral 四边形rectangle 矩形lozenge 菱形square正方形polygon多边形circle圆arc弧perimeter周长area面积diameter直径volume体积10．具体读法实例y=f(x) y is a function of x6×5＝30 six times (multiplied by) five equals (is equal to) thirty(x-y)(x+y) x minus y; x plus ythe fifth root of x squarey-10 y to the minus tenth (power)20 : 5＝16 : 4 the ratio of 20 to 5 equals the ration of 16 to 4 (20 is to 5 as 16 is to 4) e=1.6×10-19 e equals one point multiplied by ten to minus nineteenth power10-n ten to the minus none over n squareone over one minus n times z reversef(x)=ax2+bx+c the function of x equals a times the square of x plus b times x plus c |a|=b the absolute value of a equals that of bmax f(x) the maximum value of f(x)min f(x) the minimum value of f(x)∞ a sub n approaches / tends to infinity◊anthe limit of Sn as n gets arbitrarily large is one thirdx to the fifth power plus A over (divided by) the quantity x squared plus B, to the two-thirds power (A+B)C the quantity A plus B times CA+B=C A plus B equals CA-B=C A minus B equals CA×B=C A multiplied by B equals CA/B=C A divided by B equals CA : B=C : D A is toB of A to B asC is to D11．数学问题求解的一般表示Solve the following system of equationsSolution: multiply equation (1) by (2) and getSubtract equation (2) from equation (4), and getSubtract equation (3) from equation (2), and getfrom equation (5) from equation (6),obtain x and y.。
solutions to complex analysis stein.复分析部分答案
semi-circle.
Solution of Problem 3. For R > 0 let CR be the upper half-circle of radius R centered at the origin in the counter-clockwise sense. For 0 < r < R we
θ=
π 6
θ=0
π
π
≤
4
e−
1 2
R2 Rdθ
+
6 e−R2θRdθ
θ=
π 6
θ=0
=
π
4
e−
1 2
R2 Rdθ
+
θ=
π 6
−
1 R2
e−R2
θ
R
π 6
θ=0
which approaches 0 as R → ∞.
Next we observe that
∞ e−x2dx = √π,
x=−∞
because
x=0
√
dx =
2π 4
.
These are the Fresnel integrals. Here,
∞ 0
is
interpreted
as
limR→∞
0R.
Hint: Integrate the holomorphic function f (z) = e−z2 over the path which is
∞
= 2i
t=0
1 r+a
r−a
r+a
1+
r−a r+a
t
2
+
英文的数论题小书3解答
GlossaryArithmetic functionA function defined on the positive integers that is complex valued.Arithmetic-Geometric Means InequalityIf n is a positive integer and a1,a2,...,a n are nonnegative real numbers,then1 nni=1a i≥(a1a2···a n)1/n,with equality if and only if a1=a2=···=a n.This inequality is a special case of the power mean inequality.Base-b representationLet b be an integer greater than1.For any integer n≥1there is a unique system (k,a0,a1,...,a k)of integers such that0≤a i≤b−1,i=0,1,...,k,a k=0 andn=a k b k+a k−1b k−1+···+a1b+a0.Beatty’s theoremLetαandβbe two positive irrational real numbers such that1α+1β=1.The sets{ α , 2α , 3α ,...},{ β , 2β , 3β ,...}form a partition of the set of positive integers.190104Number Theory ProblemsBernoulli’s inequalityFor x>−1and a>1,(1+x)a≥1+ax,with equality when x=0.B´e zout’s identityFor positive integers m and n,there exist integers x and y such that mx+by= gcd(m,n).Binomial coefficientn k=n!k!(n−k)!,the coefficient of x k in the expansion of(x+1)n.Binomial theoremThe expansion(x+y)n=nx n+n1x n−1y+n2x n−2y+···+nn−1xy n−1+nny n.Canonical factorizationAny integer n>1can be written uniquely in the formn=pα11···pαk k,where p1,...,p k are distinct primes andα1,...,αk are positive integers. Carmichael numbersThe composite integers n satisfying a n≡a(mod n)for every integer a. Complete set of residue classes modulo nA set S of integers such that for each0≤i≤n−1there is an element s∈S with i≡s(mod n).Glossary191Congruence relationLet a,b,and m be integers,with m=0.We say that a and b are congruent modulo m if m|(a−b).We denote this by a≡b(mod m).The relation“≡”on the set Z of integers is called the congruence relation.Division algorithmFor any positive integers a and b there exists a unique pair(q,r)of nonnegative integers such that b=aq+r and r<a.Euclidean algorithmRepeated application of the division algorithm:m=nq1+r1,1≤r1<n,n=r1q2+r2,1≤r2<r1,...r k−2=r k−1q k+r k,1≤r k<r k−1,r k−1=r k q k+1+r k+1,r k+1=0This chain of equalities isfinite because n>r1>r2>···>r k.Euler’s theoremLet a and m be relatively prime positive integers.Thenaϕ(m)≡1(mod m).Euler’s totient functionThe functionϕ(m)is defined to be the number of integers between1and n that are relatively prime to m.Factorial base expansionEvery positive integer k has a unique expansionk=1!·f1+2!·f2+3!·f3+···+m!·f m,where each f i is an integer,0≤f i≤i,and f m>0.192104Number Theory ProblemsFermat’s little theoremLet a be a positive integer and let p be a prime.Thena p≡a(mod p).Fermat numbersThe integers f n=22n+1,n≥0.Fibonacci sequenceThe sequence defined by F0=1,F1=1,and F n+1=F n+F n−1for every positive integer n.Floor functionFor a real number x there is a unique integer n such that n≤x<n+1.We say that n is the greatest integer less than or equal to x or thefloor of x and we write n= x .Fractional partThe difference x− x is called the fractional part of x and is denoted by{x}.Fundamental theorem of arithmeticAny integer n greater than1has a unique representation(up to a permutation)as a product of primes.Hermite’s identityFor any real number x and for any positive integer n,x ++1n++2n+···++n−1n= nx .Legendre’s formulaFor any prime p and any positive integer n,e p(n)=i≥1np i.Glossary 193Legendre’s functionLet p be a prime.For any positive integer n ,let e p (n )be the exponent of p in the prime factorization of n !.Linear Diophantine equationAn equation of the forma 1x 1+···+a n x n =b ,where a 1,a 2,...,a n ,b are fixed integers.Mersenne numbersThe integers M n =2n −1,n ≥1.M¨o bius functionThe arithmetic function µdefined by µ(n )=⎧⎨⎩1if n =1,0if p 2|n for some prime p >1,(−1)k if n =p 1···p k ,where p 1,...,p k are distinct primes .M¨o bius inversion formulaLet f be an arithmetic function and let F be its summation function.Then f (n )= d |nµ(d )F n d .Multiplicative functionAn arithmetic function f =0with the property that for any relatively prime positive integers m and n ,f (mn )=f (m )f (n ).Number of divisorsFor a positive integer n denote by τ(n )the number of its divisors.It is clear that τ(n )= d |n1.194104Number Theory ProblemsOrder modulo mWe say that a has order d modulo m,denoted by ord m(a)=d,if d is the smallest positive integer such that a d≡1(mod m).Perfect numberAn integer n≥2with the property that the sum of its divisors is equal to2n. Pigeonhole PrincipleIf n objects are distributed among k<n boxes,some box contains at least two objects.Prime number theoremThe relationlim n→∞π(n)n/log n=1,whereπ(n)denotes the number of primes less than or equal to n.Prime number theorem for arithmetic progressionsFor relatively prime integers a and r,letπa,d(n)denote the number of primes in the arithmetic progression a,a+d,a+2d,a+3d,...that are less than or equal to n.Thenlim n→∞πa,d(n)n/log n=1ϕ(d).This result was conjectured by Legendre and Dirichlet and proved by Charles De la Vall´e e Poussin.Sum of divisorsFor a positive integer n denote byσ(n)the sum of its positive divisors including 1and n itself.It is clear thatσ(n)=d|nd.Glossary195Summation functionFor an arithmetic function f the function F defined byF(n)=d|nf(d).Wilson’s theoremFor any prime p,(p−1)!≡−1(mod p).Zeckendorf representationEach nonnegative integer n can be written uniquely in the formn=∞k=0αk F k,whereαk∈{0,1}and(αk,αk+1)=(1,1)for each k.Further Reading1.Andreescu,T.;Feng,Z.,101Problems in Algebra from the Training of theUSA IMO Team,Australian Mathematics Trust,2001.2.Andreescu,T.;Feng,Z.,102Combinatorial Problems from the Training ofthe USA IMO Team,Birkh¨a user,2002.3.Andreescu,T.;Feng,Z.,103Trigonometry Problems from the Training ofthe USA IMO Team,Birkh¨a user,2004.4.Andreescu,T.;Feng,Z.,A Path to Combinatorics for Undergraduate Stu-dents:Counting Strategies,Birkh¨a user,2003.5.Feng,Z.;Rousseau,C.;Wood,M.,USA and International MathematicalOlympiads2005,Mathematical Association of America,2006.6.Andreescu,T.;Feng,Z.;Loh,P.,USA and International MathematicalOlympiads2004,Mathematical Association of America,2005.7.Andreescu,T.;Feng,Z.,USA and International Mathematical Olympiads2003,Mathematical Association of America,2004.8.Andreescu,T.;Feng,Z.,USA and International Mathematical Olympiads2002,Mathematical Association of America,2003.9.Andreescu,T.;Feng,Z.,USA and International Mathematical Olympiads2001,Mathematical Association of America,2002.10.Andreescu,T.;Feng,Z.,USA and International Mathematical Olympiads2000,Mathematical Association of America,2001.11.Andreescu,T.;Feng,Z.;Lee,G.;Loh,P.,Mathematical Olympiads:Prob-lems and Solutions from Around the World,2001–2002,Mathematical As-sociation of America,2004.198104Number Theory Problems12.Andreescu,T.;Feng,Z.;Lee,G.,Mathematical Olympiads:Problems andSolutions from Around the World,2000–2001,Mathematical Association of America,2003.13.Andreescu,T.;Feng,Z.,Mathematical Olympiads:Problems and Solutionsfrom Around the World,1999–2000,Mathematical Association of America, 2002.14.Andreescu,T.;Feng,Z.,Mathematical Olympiads:Problems and Solutionsfrom Around the World,1998–1999,Mathematical Association of America, 2000.15.Andreescu,T.;Kedlaya,K.,Mathematical Contests1997–1998:OlympiadProblems from Around the World,with Solutions,American Mathematics Competitions,1999.16.Andreescu,T.;Kedlaya,K.,Mathematical Contests1996–1997:OlympiadProblems from Around the World,with Solutions,American Mathematics Competitions,1998.17.Andreescu,T.;Kedlaya,K.;Zeitz,P.,Mathematical Contests1995–1996:Olympiad Problems from Around the World,with Solutions,American Mathematics Competitions,1997.18.Andreescu,T.;Enescu,B.,Mathematical Olympiad Treasures,Birkh¨a user,2003.19.Andreescu,T.;Gelca,R.,Mathematical Olympiad Challenges,Birkh¨a user,2000.20.Andreescu,T.,Andrica,D.,An Introduction to Diophantine Equations,GILPublishing House,2002.21.Andreescu,T.;Andrica,D.,360Problems for Mathematical Contests,GILPublishing House,2003.22.Andreescu,T.;Andrica,D.,Complex Numbers from A to Z,Birkh¨a user,2004.23.Beckenbach,E.F.;Bellman,R.,An Introduction to Inequalities,New Math-ematical Library,V ol.3,Mathematical Association of America,1961. 24.Coxeter,H.S.M.;Greitzer,S.L.,Geometry Revisited,New MathematicalLibrary,V ol.19,Mathematical Association of America,1967.25.Coxeter,H.S.M.,Non-Euclidean Geometry,The Mathematical Associa-tion of America,1998.Further Reading199 26.Doob,M.,The Canadian Mathematical Olympiad1969–1993,Universityof Toronto Press,1993.27.Engel,A.,Problem-Solving Strategies,Problem Books in Mathematics,Springer,1998.28.Fomin,D.;Kirichenko,A.,Leningrad Mathematical Olympiads1987–1991,MathPro Press,1994.29.Fomin,D.;Genkin,S.;Itenberg,I.,Mathematical Circles,American Math-ematical Society,1996.30.Graham,R.L.;Knuth, D.E.;Patashnik,O.,Concrete Mathematics,Addison-Wesley,1989.31.Gillman,R.,A Friendly Mathematics Competition,The Mathematical As-sociation of America,2003.32.Greitzer,S.L.,International Mathematical Olympiads,1959–1977,NewMathematical Library,V ol.27,Mathematical Association of America, 1978.33.Holton,D.,Let’s Solve Some Math Problems,A Canadian MathematicsCompetition Publication,1993.34.Kazarinoff,N.D.,Geometric Inequalities,New Mathematical Library,V ol.4,Random House,1961.35.Kedlaya,K;Poonen,B.;Vakil,R.,The William Lowell Putnam Mathemat-ical Competition1985–2000,The Mathematical Association of America, 2002.36.Klamkin,M.,International Mathematical Olympiads,1978–1985,NewMathematical Library,V ol.31,Mathematical Association of America, 1986.37.Klamkin,M.,USA Mathematical Olympiads,1972–1986,New Mathemat-ical Library,V ol.33,Mathematical Association of America,1988.38.K¨u rsch´a k,J.,Hungarian Problem Book,volumes I&II,New MathematicalLibrary,V ols.11&12,Mathematical Association of America,1967. 39.Kuczma,M.,144Problems of the Austrian–Polish Mathematics Competi-tion1978–1993,The Academic Distribution Center,1994.40.Kuczma,M.,International Mathematical Olympiads1986–1999,Mathe-matical Association of America,2003.200104Number Theory Problemsrson,L.C.,Problem-Solving Through Problems,Springer-Verlag,1983.usch,H.The Asian Pacific Mathematics Olympiad1989–1993,Aus-tralian Mathematics Trust,1994.43.Liu,A.,Chinese Mathematics Competitions and Olympiads1981–1993,Australian Mathematics Trust,1998.44.Liu,A.,Hungarian Problem Book III,New Mathematical Library,V ol.42,Mathematical Association of America,2001.45.Lozansky,E.;Rousseau,C.Winning Solutions,Springer,1996.46.Mitrinovic,D.S.;Pecaric,J.E.;V olonec,V.Recent Advances in GeometricInequalities,Kluwer Academic Publisher,1989.47.Mordell,L.J.,Diophantine Equations,Academic Press,London and NewYork,1969.48.Niven,I.,Zuckerman,H.S.,Montgomery,H.L.,An Introduction to the The-ory of Numbers,Fifth Edition,John Wiley&Sons,Inc.,New York,Chich-ester,Brisbane,Toronto,Singapore,1991.49.Savchev,S.;Andreescu,T.Mathematical Miniatures,Anneli Lax NewMathematical Library,V ol.43,Mathematical Association of America, 2002.50.Sharygin,I.F.,Problems in Plane Geometry,Mir,Moscow,1988.51.Sharygin,I.F.,Problems in Solid Geometry,Mir,Moscow,1986.52.Shklarsky,D.O;Chentzov,N.N;Yaglom,I.M.,The USSR Olympiad Prob-lem Book,Freeman,1962.53.Slinko,A.,USSR Mathematical Olympiads1989–1992,Australian Mathe-matics Trust,1997.54.Szekely,G.J.,Contests in Higher Mathematics,Springer-Verlag,1996.55.Tattersall,J.J.,Elementary Number Theory in Nine Chapters,CambridgeUniversity Press,1999.56.Taylor,P.J.,Tournament of Towns1980–1984,Australian MathematicsTrust,1993.57.Taylor,P.J.,Tournament of Towns1984–1989,Australian MathematicsTrust,1992.Further Reading201 58.Taylor,P.J.,Tournament of Towns1989–1993,Australian MathematicsTrust,1994.59.Taylor,P.J.;Storozhev,A.,Tournament of Towns1993–1997,AustralianMathematics Trust,1998.60.Yaglom,I.M.,Geometric Transformations,New Mathematical Library,V ol.8,Random House,1962.61.Yaglom,I.M.,Geometric Transformations II,New Mathematical Library,V ol.21,Random House,1968.62.Yaglom,I.M.,Geometric Transformations III,New Mathematical Library,V ol.24,Random House,1973.Indexarithmetic functions,36base-b representation,41 Beatty’s theorem,60 Bernoulli’s inequality,145B´e zout’s identity,13binomial theorem,5canonical factorization,8 Carmichael numbers,32 ceiling,52Chinese remainder theorem,22 complete set of residue classes,24 composite,5congruence relation,19 coprime,11decimal representation,41 Diophantine equations,14 division algorithm,4Euclidean algorithm,12Euler’s theorem,28Euler’s totient function,27 factorial base expansion,45 Fermat numbers,22,70 Fermat’s little theorem,28 Fibonaccinumbers,45sequence,45fractional part,52fully divides,9fundamental theorem ofarithmetic,7geometric progression,9greatest common divisor,11Hermite identity,63inverse of a modulo m,26least common multiple,16of a1,a2,...,a n,16Legendre function,65linear combinations,14linear congruence equation,22linear congruence system,22linear Diophantine equation,38Mersenne numbers,71multiplicative arithmetic functions,18M¨o bius function,36M¨o bius inversion formula,37number of divisors,17order d modulo m,32perfect cube,2perfect numbers,72perfect power,2perfect square,2pigeonhole principle,93prime,5203204Indexprime number,5quotient,4reduced complete set of residueclasses,28 relatively prime,11 remainder,4square free,2sum of positive divisors,18 summation function,36twin primes,6Wilson’s theorem,26 Wolstenholme’s theorem,115 Zeckendorf representation,45。
Intro_complex_number
w s͑cos ϩ i sin ͒ and z r ͑cos ϩ i sin ͒ and using De Moivre’s Theorem, we get
s n͑cos n ϩ i sin n͒ r ͑cos ϩ i sin ͒
Im
r
¨
0
a
FIGURE 4
a+bi b
Re
Im
1+i
œ„2
π 4
0
_
π 6
Re
2
œ3„-i
FIGURE 5
Polar Form
We know that any complex number z a ϩ bi can be considered as a point ͑a, b͒ and that any such point can be represented by polar coordinates ͑r, ͒ with r ജ 0. In fact,
Re
z1 z2
r1 r2
͓cos͑1
Ϫ
2͒
ϩ
i sin͑1
Ϫ
2 ͔͒
z2 0
z¡z™ FIGURE 6
Im
r
¨
0 _¨
1
1
r
z
FIGURE 7
z Re
Im z=1+i zw
œ„2
2 œ„2
π
12
0
Re
2 w=œ„3-i
FIGURE 8
In particular, taking z1 1 and z2 z (and therefore 1 0 and 2 ), we have the following, which is illustrated in Figure 7.
SAT考试备考数学题目41. Complex numbers
Correct answer: C Difficulty level: 2
2. 4 Which of the following is equivalent to the complex number shown above? Note: √ 1 A. 15+8i B. 15−8i C. 17+8i D. 17−8i
16. 3 Which of the following is equivalent to the complex number shown above? Note: √ 1 A. 8−26i B. 18−26i C. 27−26i D. 30−26i
Correct answer: B Difficulty level: 3
Correct answer: A Difficulty level: 2
4. (5+i)(7−3i) Which of the following is equivalent to the complex number shown above? Note: √ 1 A. 32+8i B. 32−8i C. 38+8i D. 38−8i
Correct answer: C Difficulty level: 3
30. 6
22
Which of the following is equivalent to the complex number shown above?
Note: √ 1
A. 6−8i
B. 8−8i
C. 10−8i
D. 12−8i
7. 6 2 Which of the following is equivalent to the complex number shown above? Note: √ 1 A. 40 4 B. 40 24 C. 32 24 D. 32 4
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
Advances in Theoretical and Applied MathematicsISSN0973-4554V ol.1No.1(2006),pp.91-95c Research India Publications/atam.htmComplex Integer Solutions of x,y nu,v(n=1,5)M.A.Gopalan1and S.Devibala21Department of Mathematics,National College,Trichy,India.E-mail:gopalanma@2Department of Mathematics,Shrimati Indira Gandhi College,Trichy,IndiaE-mail:devibala27@AbstractDifferent patterns of two parametric solutions in complex integers of the pair ofsimultaneous equations x+y=u+v,x5+y5=u5+v5are obtained.A fewinteresting relations among the solutions are also given.AMS Mathematics Subject Classification:11D41.Keywords:Double equations,Gaussian integers,fifth order Ramanujan numbers.1.IntroductionThe notation A1,A2,...,A p n1n2,...,n r B1,B2,...,B q designates a so called multi-degreed equality and means that the sum of the numbers on the left equals the sum of the numbers on the right for each of the r(n1,n2,...,n r)positive integral powers of the numbers.In[1-5],the problems of determining two sets of non-zero integers(x1,x2,x3) and(X1,X2,X3)satisfying the system of equations x1,x2,x31,2X1,X2,X3,x1,x2,x3 2,4X1,X2,X3,x1,x2,x31,3X1,X2,X3respectively are obtained.In this communication,the system of double equations given by x,y n u,v(n=1,5) is considered and different varieties of treatment have been given to this equation.It is worth mentioning here that the solutions of the above system exhibit the existence of infinitely many double coincidences among thefifth order Ramanujan numbers in the complexfield.Such a relation seems to be not existing with integral solutions in the real field[6,7].Various relationships between variables in the equations are found.92M.A.Gopalan and S.Devibala2.Method of AnalysisThe system of equation to be solved isx+y=u+v,x5+y5=u5+v5(1,2) The choicex=p+q,y=p−q,u=p+s,v=p−s(q=s)(3) automatically satisfies(1).The substitution of(3)in(2)gives(iq)2+(is)2=2p2(4) This equation reveals the fact that there are no real values for q and s satisfying(4). Now,we present different methods offinding solutions of(4)in complex integers.Method IThe assumption(iq)=α+iβ,(is)=α−iβ(5) in(4)givesα2=β2+p2(6) which is satisfied byα=A2+B2,β=2AB,p=A2−B2(A>B>0)(7) Using(7)in(5),we getq=2AB−i(A2+B2),s=−2AB−i(A2+B2)Employing(3),the solutions of(1,2)in complex integers are found to be x=A2−B2+2AB−i(A2+B2),y=A2−B2−2AB+i(A2+B2)u=A2−B2−2AB−i(A2+B2),v=A2−B2+2AB+i(A2+B2)A few numerical examples are given below:A B x y u v217−i5−1+i5−1−i57+i54228−i20−4+i20−4−i2028+i203114−i102+i102−i1014+i103217−i13−7+i13−7−i1317+i13 An immediate result is x,v and y,u are complex conjugates respectively.If x i(= x1i+ix2i),y i(=y1i+iy2i),u i(=u1i+iu2i),v i(=v1i+iv2i)represent the solutions of (1,2),corresponding to(A i,B i)then the solutions corresponding to either(A i+1,B i+ 1)or(A i+1,1)are given by2[x1i+ix2i,−y1i+iy2i,−u1i+iu2i,v1i+iv2i],whereas corresponding to(1,B i+1)the solutions are given by2[−x1i+ix2i,y1i+iy2i,u1i+iu2i,−v1i+iv2i].Complex Integer Solutions of x,y n u,v(n=1,5)93Method2Settingq=α+iβ,s=α−iβ(8) in(4),we getβ2=α2+p2(9) whose solution isβ=A2+B2,α=A2−B2,p=2AB(10) Using(11)in(9),we obtainq=(A2−B2)+i(A2+B2),s=A2−B2−i(A2+B2) Using(3),the corresponding solutions of(1,2)are obtained asx=2AB+A2−B2+i(A2+B2),y=2AB−A2+B2−i(A2+B2)u=2AB+A2−B2−i(A2+B2),v=2AB−A2+B2+i(A2+B2) ExamplesA B x y u v217+i51−i57−i51+i54228+i204−i2028−i204+i203114+i10−2−i1014−i10−2+i103217+i137−i1317−i137+i13Note that x,u and y,v are complex conjugates respectively.The solutions corre-sponding to either(A i+1,B i+1)or(A i+1,1)are given by2[x1i+ix2i,−y1i+iy2i,u1i+iu2i,−v1i+iv2i],whereas corresponding to(1,B i+1)the solutions are given by2[−x1i+ix2i,y1i+iy2i,−u1i+iu2i,v1i+iv2i],Method3Takingq=i(α+β),s=i(α−β)(11) in(4),we obtainp2=α2+β2(12) in which we takep=A2+B2,α=2AB,β=A2−B2(13)94M.A.Gopalan and S.Devibala From(12)and(14),we haveq=i(2AB+A2+B2),s=i(2AB−A2+B2)Thus from(3),the complex integer solutions of(1,2)are seen to bex=A2+B2+i(2AB+A2−B2),y=A2+B2−i(2AB+A2−B2)u=A2+B2+i(2AB−A2+B2),v=A2+B2−i(2AB−A2+B2) ExamplesA B x y u v215+i75−i75+i5−i4220+i2820−i2820+i420−i43110+i1410−i1410−i210+i23213+i1713−i1713+i713−i7We observe that x,y and u,v are complex conjugates respectively.The solutions corresponding to either(A i+1,B i+1)or(A i+1,1)are given by2[x1i+ix2i,y1i+iy2i,u1i−iu2i,v1i−iv2i],whereas corresponding to(1,B i+1)the solutions are given by2[x1i−ix2i,y1i−iy2i,u1i+iu2i,v1i+iv2i],The above three patterns of solutions satisfy the following relations:(i)xy+uv=(u+v)2=(x+y)2(ii)xy−uv=(v−x)(v−y)=(v−y)(y−u)=(x−u)(y−u)=(x−u)(v−x) (iii)xv−yu=(v−y)(x+y)=(v−y)(v+u)=(x+y)(x−u)=(u+v)(x−u) (iv)xv+yu=−(v−x)2=−(y−u)2(v)xu+yv=(u+v)2+(v−x)2=(u+v)2+(y−u)2=(x+y)2+(v−x)2= (x+y)2+(y−u)2(vi)2(xv+yu)=(u+v)2+(x−u)2−(v−x)2=(u+v)2+(x−u)2−(y−u)2= (u+v)2+(v−y)2−(v−x)2=(u+v)2+(v−y)2−(y−u)2=(x+y)2+(x−u)2−(v−x)2=(x+y)2+(x−u)2−(y−u)2=(x+y)2+(v−y)2−(v−x)2= (x+y)2+(v−y)2−(y−u)2(vii)yv−xu=(u+v)(v−x)=(u+v)(y−u)=(x+y)(v−x)=(x+y)(y−u) (viii)The solutions corresponding to k(A i,B i)are given by k2(x i,y i,u i,v i)Complex Integer Solutions of x,y n u,v(n=1,5)95RemarkLet N1,N2be two integers written as the sum of5th powers of two Gaussian integers in two different ways.Then,the product N1N2leads to different patterns of multivariate equalities.For example,ConsiderN1=(5+i7)5+(5−i7)5=(5+i)5+(5−i)5N2=(7−i5)5+(1+i5)5=(7+i5)5+(1−i5)5Therefore from the product N1N2,we have the following patterns of multivariable equalities:Sum of the5th powers of3numbers in2different ways1.(70+i24)5+(−30+i32)5+(−i74)5=(30+i32)5+(10−i24)5+(−i26)52.(70−i24)5+(−30−i32)5+(i74)5=(30−i32)5+(10+i24)5+(i26)5Sum of the5th powers of4Gaussian integers in2different ways3.(70+i24)5+(40+i18)5+(−30+i32)5+(−i74)5=(40−i18)5+(30−i32)5+(10+i24)5+(i26)54.(70−i24)5+(40−i18)5+(−30−i32)5+(i74)5=(40+i18)5+(30+i32)5+(10−i24)5+(−i26)5References[1]Randy,L.Ekl,1996,“Equal sums of four seventh powers,”Mathematics of Com-putation,65(216),pp.1755–1756.[2]Lander,L.J.,Parkin,T.R.,and Selfridge,J.L.,1967,“A Survey of Equal Sums ofLike Powers,”Mathematics of Computation,21(99),pp.446–459.[3]Sinha,T.N.,1971,“Geometric Methods of Solving Diophantine Equations WithEqual Sums of Like Powers,”The Mathematics Student,XXXIX(4),pp.401–405.[4]Choudhary,A.,1992,“Multigrade Chains with Odd Exponents,”Journal of Num-ber Theory,42,pp.134–140.[5]Choudhary,A.,1997,“On Equal Sums offifth powers,”Indian J.Pure Appl.Math.,28(11),pp.1443–1450.[6]Hardy,G.H and Wright,E.M.,1979,“Introduction to the Theory of Numbers,”Oxford Clarendon Press,Fifth Edition,pp.333–334.[7]Rodenstiel,E.,Dardis,J.A.,and Rosenstiel,C.R.,1991,“The Four Least Solutionsin Distinct Positive Integers of the Diophantine Equations s=x3+y3=z3+w3= u3+v3=m3+n3,”IMA Bulletin,27,pp.155–157.。