# Integral coefficients

• Jun 27th 2010, 03:31 PM
dwsmith
Integral coefficients
Let $f(x)$ be a polynomial with integral coefficients and $a\equiv b \ \mbox{(mod m)}$. Then $f(a)\equiv f(b) \ \mbox{(mod m)}$.

It make since that if a, b are in the domain, then the co-domain will also be congruent of the same integers but how do I go about doing this?
• Jun 27th 2010, 04:23 PM
chiph588@
Quote:

Originally Posted by dwsmith
Let $f(x)$ be a polynomial with integral coefficients and $a\equiv b \ \mbox{(mod m)}$. Then $f(a)\equiv f(b) \ \mbox{(mod m)}$.

It make since that if a, b are in the domain, then the co-domain will also be congruent of the same integers but how do I go about doing this?

This is because given $a\equiv b\bmod{m}$ and $c\equiv d\bmod{m}$ then $aj\equiv bj\bmod{m},\; a^k\equiv b^k\bmod{m},\; a+c\equiv b+d\bmod{m}$.

Since a polynomial is a composition of those three operations, we're done.
• Jun 27th 2010, 04:28 PM
dwsmith
I don't actually have to show anything to do with the polynomials? It is ok to say since
Quote:

Originally Posted by chiph588@
This is because given $a\equiv b\bmod{m}$ and $c\equiv d\bmod{m}$ then $aj\equiv bj\bmod{m},\; a^k\equiv b^k\bmod{m},\; a+c\equiv b+d\bmod{m}$.

Since a polynomial is a composition of those three operations, we're done.

• Jun 27th 2010, 04:36 PM
chiph588@
Quote:

Originally Posted by dwsmith
I don't actually have to show anything to do with the polynomials? It is ok to say since

I think it's enough work, but some might not.

I'll walk you through an example: $f(x)=2x^2+3x+4$

$a\equiv b\bmod{m}\implies a^2\equiv b^2\bmod{m} \implies 2a^2\equiv2b^2\bmod{m}$ (1)

$a\equiv b\bmod{m}\implies 3a\equiv3b\bmod{m} \implies 2a^2+3a\equiv2b^2+3b\bmod{m}$ from (1)

and finally $2a^2+3a+4\equiv2b^2+3b+4\bmod{m}$.

If you wanted to get formal with this I suppose inducting on $\text{deg}(f)$ would do the trick.