Question

The given equation $4xy-x-y=z^2$ has:

• A. three positive integer solutions
• B. one positive integer solutions
• C. two positive integer solutions
• D. no positive integer solutions

Answer 1

The correct option is D no positive integer solutions

Suppose all the solution of the given equation are positive integers

We write the equation in the equivalent form

$(4x-1)(4y-1)=4z^2+1$.
Let $p$ be a prime divisor of $4x-1$. Then
$4z^2+1\equiv 0$(mod $p$)
or
$(2z{)}^2\equiv -1$ (mod $p$).
On the other hand, Fermat's theorem yields
$(2z{)}^{p-1}\equiv 1$ (mod $p$)
hence
$(2z{)}^{p-1}\equiv (2z^2{)}^{\frac{p-1}{2}}\equiv (-1{)}^{\frac{p-1}{2}}\equiv 1$(mod $p$)
This implies that $p\equiv 1$ (mod $4$). It follows that all prime divisors of $4x1$ are congruent to $1$ modulo $4,$ hence $4x11$ (mod $4$), a contradiction.

Hence they are no positive integer solutions.