Find possible value of $g(3)$ Let $f(x)$ and $g(x)$ are quadratic polynomial with positive integral coefficients. Let the values of $f(g(x))$ and $g(f(x))$ at points $x=0$ and $x=1$ be as follows 
1)$f(g(0))=\alpha$ and $f(g(1))=31$
And 
2)$g(f(0))=5$ and $g(f(1))=\beta$ 
Then find possible value of $g(3)$
My try:
Let $f(x)=x^2+bx+c$ and $g(x)=x^2+dx+e$ for some positive integers $b, c, d, e$.
From given information we get the the following 
1)$ f(e)=\alpha$ and $g(c)=5$
And 
2)$$g(c)+g(1+b)=\beta -2c-2bc+e$$
$$f(e)+f(1+d)=31-2e-2de+c$$
But I see no way to get $g(3)$ from the information I found.  Any hints would be greatly appreciated.

Let $f(x)=ax^2+bx+c$, $g(x)=dx^2+ex+f$ with $a,b,c,d,e,f$ positive integers.
Then $f(0)=c\ge 1$.
If we had $f(0)\ge 2$, then $5=g(f(0))\ge 4d+2e+f\ge 7$, contradiction. We conclude $$c=1$$ and then $$d+e+f=5.$$
Now we know that $$31=f(g(1))=f(5)=25a+5b+1\ge 25+5+1=31,$$ so that by sharpness of the inequality necessarily $$a=b=1.$$
Meanwhile, we can express $\alpha$ and $\beta$ more nicely:
$$ \alpha=f(g(0))=f^2+f+1,\qquad \beta=g(f(1))=g(3).$$
However, $\alpha$ and $\beta$ are of no use anyway.
Instead, $d+e+f=5$ means that the only posisbilities for $(d,e,f)$ are 
$$(1,1,3), (1,2,2), (1,3,1), (2,1,2), (2,2,1), (3,1,1) $$
with the following corresponding values of $g(3)$ (and hence $\beta$):
$$ 15, 17, 19, 23, 25, 31.$$
