3

I'm trying to follow proof of the following theorem in my book.

Theorem 2.2. Let $f$ denote a polynomial with integral coefficients. If $a \equiv b \pmod{m} $ then $f(a) \equiv f(b)\pmod {m}$.

I understand the proof fine, except for the bit I've coloured below:

Proof: We can suppose $f(x) = c_nx^n + c_{n-1}x^{n-1} + ... +c_{0}$ where the $c_i$ are integers. Since $a \equiv b (\pmod m)$ we can apply Theorem 2.1, part 4, repeatedly to find $a^2 \equiv b^2, a^3 \equiv b^3, \cdots, a^n \equiv b^n \pmod m$, and then $\color{blue}{c_ja^j \equiv c_jb^j\pmod{m}}$ and finally $c_nba^n+c_{n-1}a^{n-1}+\cdots+c_0 \equiv c_nb^n+c_{n-1}b^{n-1}+\cdots+c_0$ by Theorem 2.1 part 3.

I don't understand what gives us that ${c_ja^j \equiv c_jb^j\pmod{m}}$? In fact, I'd expect that ${c_ja^j=c_jb^j\pmod{c_jm}}$ for any $c_j > 0$.

2.1 part 3 If $a \equiv b \pmod{m} $ and $c \equiv d \pmod{m}$ then $a+c \equiv b+d \pmod{m}$

2.1 part 4 If $a \equiv b \pmod{m} $ and $c \equiv d \pmod{m}$ then $ac \equiv bd \pmod{m}$.

Community
  • 1
J. Doe
  • 37

2 Answers2

5

$c_j \equiv c_j \pmod m$ and $a^j \equiv b^j \pmod m$ (from Theorem 2.1 part 4). Using Theorem 2.1 part 4 on these two gives the required result.

Amit Rajaraman
  • 3,248
1

Another way to see it, is to say the coefficients stay the same between f(a) and f(b). $a\equiv b\bmod m\implies a=mx+b$ and powering each side leads to all but one term divisible by m namely it leads to $a^n\equiv b^n\bmod m$ replacement then shows the coefficient need not change, so we can add them into the mix as well. by any number being congruent to itself, we get the desired result.

ADDENDUM:

we can also just hide the $c_j$ in the multiplier of m, after all, all multiples of $c_jm$ are multiples of m .