Why is $(1+80)^t\equiv 1+80t \pmod{25}$ correct?

Question

In a calculation process, I saw a modulo equation $$(1+80)^t\equiv 1+80t \pmod{25}.$$

I can't understand why it's correct.

Also, does the modulo equation $(1+n)^t\equiv 1+n^t\pmod k$ always hold? ($n, t, k$ are positive integers.)

Any help is appreciated.

Answer 1

The answer comes from the binomial expansion.

Indeed, note that : $$ (1+80)^t = \sum_{i=0}^t \binom tk 1^k80^{t-k} = 1 + 80t + 80^2M $$

where $M$ is a natural number. Now, $80^2$ is a multiple of $25$, so if we went modulo $25$ only $1+80t$ would remain.

In more generality :

Let $a,b,t,m$ be positive integers , and let $p$ be an integer such that $m$ divides $a^p$ or $m$ divides $b^p$. Then : $$ (a+b)^t \equiv \sum_{i=0}^{p-1} \binom ti a^ib^{t-i} \pmod m $$

The proof is exactly the same : in the binomial expansion, all terms above $p-1$ vanish mod $m$ because $m$ divides $a^p$.

Suppose $m > 1$ and either $a$ or $b= 1$. Then we reduce to your case.