<s>[INST] $a^2=ab+b^2+b+5$ has no integer solutions Prove that the equation $a^2=ab+b^2+b+5$ has no integer solutions.
My attempt:
$4a^2=4ab+4b^2+4b+20$
$4a^2-4ab+b^2 = 5b^2+4b+20$
$5(4a^2-4ab+b^2) = 5(5b^2+4b+20)$
$5(2a-b)^2=(5b+2)^2+96$
Let $2a-b = x$, $\;5b+2=y$
we get $\;5x^2-y^2=96$.
Please suggest on how to proceed.
 [/INST] $$5x^2-y^2 = 84 \implies 2x^2-y^2 \equiv 0 \pmod{3}$$
which implies $3\mid x$ and $3\mid y$. But $9\nmid 84$.

Your original equation should be equivalent to $$5x^2-y^2 = 96$$
where $x=2a-b, y=5b+2$. But the above method works fine for $96$ too.
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