Find all natural numbers $a,b$ such that $ab, 4a+b-3$ are perfect squares and $9a-4b$ is a prime number. Find all natural numbers $a,b$ such that $ab, 4a+b-3$ are perfect squares and $9a-4b$ is a prime number.
I can't find any idea for this problem :(

If $ab$ is a perfect square, then $a$ and $b$ must be perfect squares, say $a=m^2$ and $b=n^2$. It follows that
$$
9a-4b = (3m+2n)(3m-2n),
$$
so that for $9a-4b$ to be prime it must be or $3m+2n=1$ or $3m-2n=1$. Now we consider $4a+b-3\pmod{3}$: the only squares in $\mathbb{Z}/3\mathbb{Z}$ are $0$ and $1$, thus the only possibilities are $a\equiv 0\pmod{3}\wedge b\equiv 0\pmod{3}$ (which we can exclude immediately because in this case $9a-4b$ cannot be prime), $a\equiv 1\pmod{3}\wedge b\equiv 0\pmod{3}$ (which we can exclude because otherwise $3\mid 9a-4b$) and $a\equiv 0\pmod{3}\wedge b\equiv 1\pmod{3}$.
We now consider this last case. By splitting on whether $3m+2n=1$ or $3m-2n=1$ we get
$$
\begin{cases}
m\equiv 3\pmod{6}\\ n\equiv 2\pmod{3}
\end{cases}
\quad\vee\quad
\begin{cases}
m\equiv 3\pmod{6}\\ n\equiv 1\pmod{3}
\end{cases}.
$$
The first of these two makes it impossible to have $9a-4b$ prime. As for the second case, by writing $n=1+3k$ and $m=3+6h$ we can find that for $9a-4b$ to be prime we must have $k=3h+1$, from which $m=3+6h$ and $n=4+9h$. Now $4a+b-3=225h^2+216h+49$ is a perfect square if and only if $h=0$, so that the only solution to the problem are $a=9$ and $b=16$.
