There are two families of functions which satisfy the condition: (1) $f(n)= \begin{cases}0 & \text { if } n \text { is even, and } \\ \text { any perfect square } & \text { if } n \text { is odd }\end{cases}$ (2) $f(n)=n^{2}$, for every integer $n$. It is straightforward to verify that the two families of functions are indeed solutions. Now, suppose that f is any function which satisfies the condition that $f(f(a)-b)+b f(2 a)$ is a perfect square for every pair $(a, b)$ of integers. We denote this condition by $\left(^{*}\right)$. We will show that $f$ must belong to either Family (1) or Family (2). Claim 1. $f(0)=0$ and $f(n)$ is a perfect square for every integer $n$. Proof. Plugging $(a, b) \rightarrow(0, f(0))$ in $\left(^{*}\right)$ shows that $f(0)(f(0)+1)=z^{2}$ for some integer $z$. Thus, $(2 f(0)+1-2 z)(2 f(0)+1+2 z)=1$. Therefore, $f(0)$ is either -1 or 0 . Suppose, for sake of contradiction, that $f(0)=-1$. For any integer $a$, plugging $(a, b) \rightarrow(a, f(a))$ implies that $f(a) f(2 a)-1$ is a square. Thus, for each $a \in \mathbb{Z}$, there exists $x \in \mathbb{Z}$ such that $f(a) f(2 a)=x^{2}+1$ This implies that any prime divisor of $f(a)$ is either 2 or is congruent to $1(\bmod 4)$, and that $4 \nmid f(a)$, for every $a \in \mathbb{Z}$. Plugging $(a, b) \rightarrow(0,3)$ in $\left(^{*}\right)$ shows that $f(-4)-3$ is a square. Thus, there is $y \in \mathbb{Z}$ such that $f(-4)=y^{2}+3$. Since $4 \nmid f(-4)$, we note that $f(-4)$ is a positive integer congruent to $3(\bmod 4)$, but any prime dividing $f(-4)$ is either 2 or is congruent to $1(\bmod 4)$. This gives a contradiction. Therefore, $f(0)$ must be 0 . For every integer $n$, plugging $(a, b) \rightarrow(0,-n)$ in $\left(^{*}\right)$ shows that $f(n)$ is a square. Replacing $b$ with $f(a)-b$, we find that for all integers $a$ and $b$, $$\begin{equation*} f(b)+(f(a)-b) f(2 a) \text { is a square. } \tag{**} \end{equation*}$$ Now, let $S$ be the set of all integers $n$ such that $f(n)=0$. We have two cases: - Case 1: $S$ is unbounded from above. We claim that $f(2 n)=0$ for any integer $n$. Fix some integer $n$, and let $k \in S$ with $k>f(n)$. Then, plugging $(a, b) \mapsto(n, k)$ in $\left({ }^{* *}\right)$ gives us that $f(k)+(f(n)-k) f(2 n)=(f(n)-k) f(2 n)$ is a square. But $f(n)-k<0$ and $f(2 n)$ is a square by Claim 1. This is possible only if $f(2 n)=0$. In summary, $f(n)=0$ whenever $n$ is even and Claim 1 shows that $f(n)$ is a square whenever $n$ is odd. - Case 2: $S$ is bounded from above. Let $T$ be the set of all integers $n$ such that $f(n)=n^{2}$. We show that $T$ is unbounded from above. In fact, we show that $\frac{p+1}{2} \in T$ for all primes $p$ big enough. Fix a prime number $p$ big enough, and let $n=\frac{p+1}{2}$. Plugging $(a, b) \mapsto(n, 2 n)$ in $\left(^{* *}\right)$ shows us that $f(2 n)(f(n)-2 n+1)$ is a square for any integer $n$. For $p$ big enough, we have $2 n \notin S$, so $f(2 n)$ is a non-zero square. As a result, when $p$ is big enough, $f(n)$ and $f(n)-2 n+1=f(n)-p$ are both squares. Writing $f(n)=k^{2}$ and $f(n)-p=m^{2}$ for some $k, m \geq 0$, we have $$(k+m)(k-m)=k^{2}-m^{2}=p \Longrightarrow k+m=p, k-m=1 \Longrightarrow k=n, m=n-1$$ Thus, $f(n)=k^{2}=n^{2}$, giving us $n=\frac{p+1}{2} \in T$. Next, for all $k \in T$ and $n \in \mathbb{Z}$, plugging $(a, b) \mapsto(n, k)$ in $\left({ }^{* *}\right)$ shows us that $k^{2}+(f(n)-k) f(2 n)$ is a square. But that means $(2 k-f(2 n))^{2}-\left(f(2 n)^{2}-4 f(n) f(2 n)\right)=4\left(k^{2}+(f(n)-k) f(2 n)\right)$ is also a square. When $k$ is large enough, we have $\left|f(2 n)^{2}-4 f(n) f(2 n)\right|+1<|2 k-f(2 n)|$. As a result, we must have $f(2 n)^{2}=4 f(n) f(2 n)$ and thus $f(2 n) \in\{0,4 f(n)\}$ for all integers $n$. Finally, we prove that $f(n)=n^{2}$ for all integers $n$. Fix $n$, and take $k \in T$ big enough such that $2 k \notin S$. Then, we have $f(k)=k^{2}$ and $f(2 k)=4 f(k)=4 k^{2}$. Plugging $(a, b) \mapsto(k, n)$ to $\left(^{* *}\right)$ shows us that $f(n)+\left(k^{2}-n\right) 4 k^{2}=\left(2 k^{2}-n\right)^{2}+\left(f(n)-n^{2}\right)$ is a square. Since $T$ is unbounded from above, we can take $k \in T$ such that $2 k \notin S$ and also $\left|2 k^{2}-n\right|>\left|f(n)-n^{2}\right|$. This forces $f(n)=n^{2}$, giving us the second family of solution.