Consider multiple polynomials with coefficients in $\mathbb Z$ and of degree at least 2 (thanks Moos): $g_1,g_2..g_i$. How can I go about showing that there are an infinite amount of rationals number $t$ so that for any $s$ in $\mathbb Q$, $g_j(s)\ne t$ for $j=1,..i$?
I can do this for one polynomial, but not yet for many. Elementary solutions are preferred.
edit: I'd rather not a full solution but hints.
Okay:
It clearly suffices to show the existence of one such $t$, I'll even make it an integer. We're going to play around with an integer $-d$ till it works:
Let $g_1$ be $a_n x^n+...+ a_0$, then the solutions $p/q$ of $g_1(x)=-d$ satisfy $p|(a_0+d)$, and $q|a_n$. Specifically, the denominator divides $a_n$, and so has a finite amount of options. So any roots that come out of $g_1$ for integers will look like $r/a_n$ for some $r$. Let's look at the sequence:
$g_1(1/a_n)$ , $g_1(2/a_n)$, $g_1(3/a_n)$....
The difference between successive terms grows at least linearly in absolute value (an easy way to see this is by Lagranges theorem: $g_1(x+1/n)-g_1(x)=n*g_1'(x')$. The same holds if we plug in $g_1(-1/a_n)$, $g_1(-2/a_n)$...
Of course all this is true for all $g_j$, since the spaces between successive terms grows really fast the result is obtained.
I avoided a few details that are too messy to actually write: from some point the polynomial are monotone, and you can look like at the image of $g_r$ that are integers in [-n,n], use b(n) to count this. Then what I showed is b(n)/n -> 0 and therefore there are infinitely many integers not there.