Show that there are no natural numbers \(x,y\ge1\) such that \[ x^2+y^2=7xy. \]
Solution
Assume that there was such a solution. Taking remainders modulo 7 gives us
\[ x ^ { 2 } + y ^ { 2 } \equiv 0 \mod 7 . \]
The quadratic remainders modulo 7 are 0, 1, 2, 4. The only two quadratic remainders whose sum is \(\equiv 0\) are 0 and 0. So
\[ x ^ { 2 } \equiv y ^ { 2 } \equiv 0 \mod 7 . \]
It follows that \(x , y\) are both divisible by 7, i.e. \(x = 7 x _ { 1 } , y = 7 y _ { 1 }\) , for some natural numbers \(x _ { 1 } , y _ { 1 }\) . It follows that
\[ x _ { 1 } ^ { 2 } + y _ { 1 } ^ { 2 } = 7 x _ { 1 } y _ { 1 } . \]
Repeating this process would produce an infinite sequence of pairs \(( x , y ) , ( x _ { 1 } , y _ { 1 } ) , ( x _ { 2 } , y _ { 2 } ) , . . .\) such that \(x _ { i }\) and \(y _ { i }\) are strictly decreasing sequences of integers. Contradiction.