关于不定方程组y^2-(k^2+1)x^2=k^2,z^2-((k+1)~2+1)x^2=(k+1)~2

On the Coupled Indefinite Equations y^2-(k^2+1)x^2=k^2,z^2-((k+1)~2+1)x^2=(k+1)~2

证明了标题所述的不定方程组k＝7或k＝8时仅有x＝0的整数解，从而证明了不存在异于1的正整数N，使1，50，65，N和1，65，82，N的任两数之乘积减去1后均为平方数，所用的方法属于初等的。

The coupled indefinite equations y2 - (k2 + 1 )x2 = k2 and z2 - ((k + 1 )2 + 1 ) x2 = (k + 1 ) 2are proved. When k equal to 7 or 8, their integral solution is only x= 0. Which has proved that the positive integral N except 1 does not exist. In such a condition any poduct between 1,5o, 65, N and 1, 65, 82, N minus 1 is a square numbers. Such a solution can be reached by an elementary method.