Find triples (a,b,c) such that $f(n)f(n+1)=f(m)$ where $f(x)=ax^2+bx+c$ 
Find all triples of integers $(a,b,c)$ with $a\neq0$ such that the function $f(x)=ax^2+bx+c$ has the property that, for each positive integer $n$, there is an integer $m$ with $$f(n)f(n+1)=f(m).$$

Attempt: Writing$\,\,f(x)=a(x+\alpha)(x+\beta)$, we get $$f(n)f(n+1)=a^2\left(z+\alpha)(z+\beta\right)$$ where $z=n^2+n\left(1+\frac{b}{a}\right)+\frac{c}{a}$. Hence all triples $(1,b,c)$ work. Are these the only ones?
EDIT. I am able to show if $a=m^2, a\neq 1$, then the triples are exactly $(m^2,2mn,n^2)$ where $m$ divides  $n(n-1)$. So now the question is whether $a$ must be a perfect square.
P.S. This is the problem 11964 from AMM problem section (March 2017) for which the deadline for submitting solutions has ended.
 A: Negative $a$ will certainly not work: It makes $f(m)$ bounded from above whereas $f(n)f(n+1)$ is not.
Your argument is correct in showing that all triples with $a=1$ work. 
We vary the problem by demanding the desired property only for all sufficiently large $n$. Then $(a,b,c)$ has the property if and only if $(a,b\pm 2a,a\pm b+c)$ has it. Thus we may assume wlog that $|b|\le a$ (equivalently, $f$ has its minimum in $[-\frac12,\frac12]$).
Case 1: Assume $b=0$. Now the equation is 
$$ a^2n^4+2a^2n^3+a(a+c)n^2+2acn+(a+c)c = am^2+c.$$
So $m^2 = an^4+O(n^3)$, hence $m=\pm n^2\sqrt a+O(n)$.
With $m=n^2\sqrt a+u n$, 
$$ a^2n^4+2a^2n^3+a(a+c)n^2+2acn+(a+c)c= a^2n^4+2au\sqrt an^3+u^2an^2+c,$$
or,
$$ 2a\sqrt a(\sqrt a-u)= (u^2-a-c)an^{-1}-2acn^{-2}+(1-a-c)cn^{-3}$$
As $u$ is bounded, we conclude that $u=\sqrt a+O(n^{-1})$.
This makes $m=n^2\sqrt a+n\sqrt a+v$ with $v=O(1)$.
Now we arrive at
$$a(c-2v\sqrt a) = 2a(\sqrt a v-c) n^{-1} +v^2a+(1-a-c)cn^{-2}$$
and from this $v=\frac c{\sqrt a}+O(n^{-1})$, 
$$m=\sqrt a\cdot(n^2+n+\tfrac ca)+O(n^{-1}).$$
Case 1.1: If $a$ is a perfect square, this implies that $c$ is a multiple of $\sqrt a$,  and we can write $a=d^2$, $c=kd$; then for $n$ large enough, we have $m=dn^2+dn+k$ and then $f(n)f(n+1)=f(m)$ leads to $kd^3-kd$, i.e., $k=0$ or $a=1$, i.e., the solutions found by the OP and in the comments.
Case 1.2: If $a$ is not a perfect square ...
Case 2: If $b\ne 0$, ...
