Quadratic trinomial must have integer coefficient ot not $?$

Yes. In original problem all coefficients are integers.

Sorry, I have now edited it.

I have done something here, dont know if this is ok set $f(x)=ax^2+bx+c$, by quadratic formula for $f(f(x))$ $ax^2+bx+c=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ then we go down further to get $$x=\frac{1}{2a}(-b \pm \sqrt{2(-b \pm \sqrt{b^2-4ac})+\frac{b^2-4ac}{a}})$$if the square root part is not an integer, then $x^2$ wont be an integer so let that sqrt part be k again if $\frac{-b+k}{2a}$ is not an integer , then $x^2$ wont be an integer so x is integer, now find largest $x^2$ not exceeding 2015, which is $44^2= 1936$ dont know if this is correct

cyborg_w wrote: I have done something here, dont know if this is ok set $f(x)=ax^2+bx+c$, by quadratic formula for $f(f(x))$ $ax^2+bx+c=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ then we go down further to get $$x=\frac{1}{2a}(-b \pm \sqrt{2(-b \pm \sqrt{b^2-4ac})+\frac{b^2-4ac}{a}})$$if the square root part is not an integer, then $x^2$ wont be an integer so let that sqrt part be k again if $\frac{-b+k}{2a}$ is not an integer , then $x^2$ wont be an integer so x is integer, now find largest $x^2$ not exceeding 2015, which is $44^2= 1936$ dont know if this is correct I think this is right and indeed is a very good solution .

wait... i dont know how to verify that 44 is possible, pls help

For $n=44^2 f(x)=x(x-44)$ $f(f(\sqrt{n})=f(f(44))=f(0)=0$ Btw, $n=44^2$ is not maximum. For $n=1980 f(x)=x^2-2025$ $f(f(\sqrt{n}))=f(1980-2025)=f(-45)=0$

but yours is not a trinomial as a trinomial has 3 terms i suppose

In original problem $f(x)$ is quadratic polynomial with integer coeffcients. So $f(x)=x^2-2025$ is possible.

I just checked the original Russian paper, and it says trinomial lol

$f(x)=x^2-44x+43$. $f(x) = (x-1)(x-43)$ $f(f(x)) = x(x-44)(x^2-44x+42)$ $f(f(\sqrt{44^{2}}))=f(f(44)) = 0$ If $f(x)=ax^2+bx+c$ ($ab\neq0$) satisfies the condtion, then you can prove that $n$ is a perfect square. (cyborg_w has already argued it, but he needs more details because he didn't use $b\neq0$.

The right answer is $2010$ , so nobody solved this yet... I've only found an example : $-3x^2 + 6075$

Let $f(x)=a(x-b)(x-c)$ where $b,c$ are rational roots And so $a(b+c)$ and $abc$ are integers $f(\sqrt{n})$ is root of $f(x)=0$ so $f(\sqrt{n})=b$ or $c$ Let $f(\sqrt{n})=b \to a(n+bc-(b+c)\sqrt{n})=b$ $(b+c)\sqrt{n}$ should be rational, so Case 1: $n$ is square Then $n \leq 44^2=1936$ Case 2: $b+c=0$ Then $a(n-b^2)=b$ or $n=b^2+\frac{b}{a}$ and $ab^2$ is integer So $b=\frac{p}{q},a=kq^2$ Then $n=\frac{p^2}{q^2}+\frac{p}{kq^3}=\frac{p(pkq+1)}{kq^3}$ is integer But $(p,q)=(q,pkq+1)=1$ so $q=1$ $n=p^2+\frac{p}{k}$ for integer $p,k$ $p^2-|p| \leq p^2 +\frac{p}{k}\leq 2015 \to |p| \leq 45$ If $|p| \leq 44$ then $p^2+\frac{p}{k} \leq 44^2+44=1980$ If $|p|=45$ then $p^2+\frac{p}{k}=2025 \pm d$ where $d$ is divisor of $45$ As $n\leq 2015 \to n=2025-d, d \geq 10$ and minimal such $d|45$ which give maximal $n$ is $d=15$ Then $n=2010$ and $p=-45,k=3 \to b=-45,a=3,c=45$ and $f(x)=3(x-45)(x+45)=3x^2-6075$ So $n=2010$