<s>[INST] if $x \equiv y$ (mod $m$), then $ax^2 + bx + c \equiv ay^2 + by + c$ (mod $m$) By assumption $y = x + km$ ; some $k \in \mathbb{Z}$. 
w.t.s  $ay^2 + by + c = (ax^2 + bx + c$) + $k'm$ ; some $k' \in \mathbb{Z}$ 
If we substitute our assumption in for $y$ we have 
$a(x + km)^2 + b(x + km) + c$ 
= $a(x^2 + 2xkm + k^2m^2) + bx + c + (bk)m$ 
= $ax^2 + 2xkma + ak^2m^2 + bx + c + (bk)m$ 
= $ax^2 + bx + c + (2xka + k^2ma + bk)m$ 
Lastly, set $k' = 2xka + k^2ma + bk$ and we have 
$ay^2 + by + c = (ax^2 + bx + c$) + $k'm$
does this work? heavily realying just on closure of $\mathbb{Z}$ under + and $\cdot$ 
(I'm fairly new to number theory)
Also, if this has been asked please just refer me to the Paige, I couldn't finds it. I apologize in advance if it has been. 
 [/INST] Prove things once.  Then assume they are written in stone.
Prove once that if $a\equiv a'\pmod n$ and $b\equiv b'\pmod n$ then $a \pm b\equiv a' \pm b'\pmod n$.
Then prove once that if $a \equiv a' \pmod n$ and $b\equiv b'\pmod n$ then $ab \equiv a'b'\pmod n$.
Then observe once by induction that if $a \equiv a'$ then $a^m \equiv a'^m \pmod n$ for all positive powers $m$.
And observe once for any integer $d$ that $d \equiv d \pmod n$ for any integer $n$.
Then this is ... just observation.

$x \equiv y \pmod m$ so $x^2 \equiv y^2$ and as $a \equiv a \pmod m$ then $ax^2 \equiv ay^2$ and as $b \equiv b$ and $c \equiv c$ then $bx \equiv by$ and $ax^2 + bx + c \equiv ax^2 + bx + c\pmod n$.

And that IS a complete proof.
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To prove these things ONCE:
$c - c = 0$ and $0 = 0*n$ for all integers $n$ so $n|c-c$ for all $n$ so

$c \equiv c \pmod n$ for any integers $c, n$.

If $n|a$ then there is an integer $k$ so that $a = kn$ and if $n|b$ then there is a $j$ so that $b = jn$ so $a\pm b = kn \pm jn = (k\pm j)n$.  $k\pm j$ are both integers so

If $n|a$ and $n|b$ then $n| a\pm b$.

If $a\equiv a' \pmod n$ and $b\equiv b' \pmod n$ then $n|a-a'$ and $n|b-b'$ so $n|(a-a') \pm (b - b')=(a\pm b)-(a' \pm b')$ so

If $a \equiv a'\pmod n$ and $b \equiv b'\pmod n$ then $a\pm b \equiv a' \pm b'\pmod n$.

If $a\equiv a' \pmod n$ then there is a $k$ so that $a = a' + kn$ and if $b\equiv b' \pmod n$ then there is a $j$ so that $b = b' + jn$ so $ab = a'b' + ajn + bkn + jkn^2 = a'b' + n(aj + bk + jkn)$ and as $aj + bk + jkn$ is an integer then

If $a \equiv a'\pmod n$ and $b \equiv b'\pmod n$ then $ab \equiv a'b' \pmod n$.

If we have $a \equiv a' \pmod n$ and we have $a^k \equiv a'^k \pmod n$ for $k = 1$.  If we assume that for some $k$ that $a^k \equiv a'^k$ then we have $a^{k+1} = a^ka \equiv a'^ka'= a'^{k+1} \pmod n$
So by induction $a^m \equiv a'^m\pmod n$ for all positive integer powers $m$.
....
And that's it.  We have proven everything we need once.  We will NEVER need to prove them again.
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