A note on the diophantine equation x 2 + b y = c z

Let a, b, c, r be positive integers such that a ² + b ² = c ^r , min(a, b, c, r) > 1, gcd(a, b) = 1, a is even and r is odd. In this paper we prove that if b ≡ 3 (mod 4) and either b or c is an odd prime power, then the equation x ² + b ^y = c ^z has only the positive integer solution (x, y, z) = (a, 2, r) with min(y, z) > 1.     

The complete solution of a diophantine equation involving biquadrates

While parametric solutions of the diophantine equation are known for any integral value of s?2, the complete solution in integers is not known for any value of s. In this paper, we obtain the complete solution of this equation when s?13.     

Diophantine方程(a-1)x~2+f(a)=4a~n的一点注记

设a是大于1的正整数,f(a)是a的非负整系数多项式,f(1)=2rp+4,其中r是大于1的正整数,p=2~l-1是Mersenne素数.本文讨论了方程(a-1)x~2+f(a)=4a~n的正整数解(x,n)的有限性,并且证明了:当f(a)=91a+9时,该方程仅当a=5,7和25时分别有解(x,n)=(3,3),(11,3)和(3,4).     

方程a2(x)(t-τ)+a1(x)(t-τ)+a0x(t-τ)+b2(x)(t)+b1(x)(t)+b0x(t)=δ的部分解讨论

讨论了方程a2(x)(t-τ)+a1(x)(t-τ)+a0x(t-τ)+b2(x)(t)+b1(x)(t)+b0x(t)=δ的部分解.     

On the diophantine equation v(v+1)=u(u+a)(u+2a)

Let \(a\in \mathbb {N}\) . We discuss the diophantine equation $$v(v+1)=u(u+a)(u+2a) $$ and some important arithmetic properties of the associated cubic field. We also present a detailed account of the cases a=2 and a=5.     

关于指数丢番图方程(an-1)((a+s)n-1)=x2可解性的研究

运用同余,整除,Pell方程等性质,其它已知结论以及初等数论方法,研究了一类与平方数有关的指数丢番图方程的可解性问题.通过将方程的参数限定在一定的数量关系下,给出判定方程无正整数解的三个充分条件,一定程度上拓展了方程无正整数解的范围,也进一步推广了前人的研究结果.     

A Conjecture Concerning the Pure Exponential Diophantine Equation a^x ＋ b^y = c^z

Let a, b, c, r be fixed positive integers such that a^2 ＋ b^2 = c^r, min（a, b, c, r） 〉 1 and 2 r. In this paper we prove that if a ≡ 2 （mod 4）, b ≡ 3 （mod 4）, c 〉 3.10^37 and r 〉 7200, then the equation a^x ＋ b^y = c^z only has the solution （x, y, z） = （2, 2, r）.     

方程f~（n）（x）＝Af~（n－k）（（ax＋b）／（cx－a））

本文提出方程ｆ~（ｎ）（ｘ）＝Ａｆ（ｎ－ｋ）（（ａｘ＋ｂ）／（ｃｘ－ａ）），证明它是可积的．所得结论是文献［１］中结论的推广．     

关于指数Diophantine方程2~x+p~y=z~2的解的个数的上界估计

设P是一个固定的奇素数.得到了方程2~x+p~y=z~2的所有正整数解(x,y,z)的一个分类.此外,证明了:如果P≡1(mod 4)并且P≠17,那么Diophantine方程2~x+p~y=z~2的全部正整数解(x,y,z)的个数N(p)满足估计N(p)≤4.     

关于Diophantine方程(a~n-1)((a+1)~n-1)=x~2

本文研究了指数Diophantine方程(an-1)((a+1)n-1)=x2的正整数解(n,x),其中a是大于1的正整数.运用初等数论方法证明了:当a≡2或3(mod4)时,该方程无解.     

指数Diophantine方程(bn)~x+(2n)~y=((b+2)n)~z的例外解

设b是大于3的正奇数.运用初等方法讨论了方程(bn)^x+(2n)x+(2n)^y=((b+2)n)y=((b+2)n)^z适合(x,y,z)≠(1,1,1)的正整数解(x,y,z,n).证明了:i)对于任何给定的正整数N,存在无穷多个b可使该方程有满足min{x,y,z}≥N的正整数解(x,y,z,n);ii)对于任何给定的b,该方程仅有有限多组正整数解(x,y,z,n)满足y>z=x.     

Diophantine方程nx~2+2~m=y~n(英文)

设n是大于3的奇数.本文运用Y.Bilu,G.Hanrot和P.M.Voutier关于Lehmer数本原素因子存在性的新近结果,证明了方程nx~2+2~m=y~n没有适合gcd(x,y)=1且m为奇数的正整数解(x,y,m).     

关于指数丢番图方程$(a^m-1)(b^n-1)=x^2$的注记

Let a and b be fixed positive integers.In this paper,using some elementary methods,we study the diophantine equation（a~m-1）（b~n-1）= x~2.For example,we prove that if a ≡ 2（mod 6）,b ≡ 3（mod 12）,then（a~n-1）（b~m-1）= x~2 has no solutions in positive integers n,m and x.     

Sur L'equation Diophantienne (xn-1)/(x-1)=yq, III

In this paper we study the diophantine equation of the title,which was first introduced by Nagell and Ljunggren during thefirst half of the twentieth century. We describe a method whichallows us, on the one hand when n is fixed, to obtain an upperbound for q, and on the other hand when n and q are fixed, toobtain upper bounds for x and y which are far sharper than thosederived from the theory of linear forms in logarithms. We alsoshow how these bounds can be used even when they seem too largefor a straightforward enumeration of the remaining possiblevalues of x. By combining all these techniques, we are ableto solve the equation in many cases, including the case whenn has a prime divisor less than 13, or the case when n has aprime divisor which is less than or equal to 23 and distinctfrom q. 2000 Mathematical Subject Classification: primary 11D41;secondary 11J86, 11Y50.     

Mordell方程y~3=x~2+2p~4的正整数解

设p是奇素数.对于非负整数r,设U_(2r+1)=(α~(2r+1)+β~(2r+1))/2~(1/2),V_(2r+1)=(α~(2r+1)-β~(2r+1))/6~(1/2),其中α=(1+3~(1/2))/2~(1/2),β=(1-3~(1/2))/2~(1/2).运用初等数论方法证明了:方程y~3=x~2+2p~4有适合gcd(x,y)=1的正整数解(x,y)的充要条件是p=U_(2m+1),其中m是正整数.当上述条件成立时,方程仅有正整数解(x,y)=(V(2m+1)(V_(2m+1)~2-6),V_(2m+1)~2+2)适合gcd(x,y)=1.由此可知:当p10000时,方程仅有正整数解(p,x,y)=(5,9,11),(19,1265,123),(71,68675,1683)和(3691,9677201305,4541163)适合gcd(x,y)=1.     

A note on the diophantine equation x2 + 1 = dy4

In this paper we prove that the Diophantine equation as in the title has at most one integer solution if $ \in > 5 \times 10^7 $ where $ \in = u + \upsilon \sqrt d $ is the least positive solution of Pell’s equation $x^2 - dy^2 = - 1$