Solution:
Suppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:
$$\begin{aligned} & 16 a - 9 b = p \\ & a b = n^2 \\ & a + b = m^2 \end{aligned}$$
Moreover, let $d = \gcd(a, b)$ and $a = d x$, $b = d y$ for some relatively prime integers $x$ and $y$. Obviously $a \neq 0$ and $b \neq 0$, $a$ and $b$ are positive (by (2) and (3)).
From (2) follows that $x$ and $y$ are perfect squares, say $x = l^2$ and $y = s^2$.
From (1), $d \mid p$ and hence $d = p$ or $d = 1$. If $d = p$, then $16 x - 9 y = 1$, and we obtain $x = 9 k + 4$, $y = 16 k + 7$ for some nonnegative integer $k$. But then $s^2 = y \equiv 3 \pmod{4}$, which is a contradiction.
If $d = 1$ then $16 l^2 - 9 s^2 = p \Rightarrow (4 l - 3 s)(4 l + 3 s) = p \Rightarrow (4 l + 3 s = p \wedge 4 l - 3 s = 1)$.
By adding the last two equations we get $8 l = p + 1$ and by subtracting them we get $6 s = p - 1$. Therefore $p = 24 t + 7$ for some integer $t$ and $a = (3 t + 1)^2$ and $b = (4 t + 1)^2$ satisfy the conditions (1) and (2). By (3) we have $m^2 = (3 t + 1)^2 + (4 t + 1)^2 = 25 t^2 + 14 t + 2$, or equivalently $25 m^2 = (25 t + 7)^2 + 1$.
Since the difference between two nonzero perfect squares cannot be $1$, we have a contradiction. As a result there is no solution.