If $a_i \equiv b_i \pmod m, i=1,2,\ldots,k$, then $a_1\cdots a_k \equiv b_1\cdots b_k \pmod m$.

Question

Let $a,b,a_i,b_i,i=1,2,\ldots,k$, and $m$ be integers with $k \ge 1$ and $m \ne 0$.

Show that: If $a_i \equiv b_i \pmod m, i=1,2,\ldots,k$, then $a_1\cdots a_k \equiv b_1\cdots b_k \pmod m$. In particular, if $a \equiv b \pmod m$, then $a^k \equiv b^k \pmod m$ for all $k$.

Attempt:

Fact. For any integers $a,b,c,d,$ and $m$ with $m \ne 0$, if $a \equiv b \pmod m$ and $c \equiv d \pmod m$, then $ac \equiv bd \pmod m$.

Proof of Fact: Let $a,b,c,d,$ and $m$ be integers with $m \ne 0$. Since $a \equiv b \pmod m$, then $a-b=ms$ for some integer $s$. Similarly, since $c \equiv d \pmod m$, then $c-d=mt$ for some integer $t$. Now, notice that $$ac-bd=ac-bc+bc-cd=c(ms)+c(mt)=m(cs+ct)m.$$ Hence, $m \mid (ac-bd)$. Thus, $ac \equiv bd \pmod m. \quad \Box$.

Proof by induction for the first part:

Base step: for $k=1$, if $a_1 \equiv b_1 \pmod m$, then $a_1 \equiv b_1 \pmod m$.
Inductive step: Assume that for $k=n$, the statement is true, i.e., if $a_i \equiv b_i \pmod m,i=1,2,\ldots,n$, then $a_1\cdots a_n \equiv b_1\cdots b_n$. We'll show that the statement is also true for $k=n+1$, i.e., if $a_i \equiv b_i \pmod m,i=1,2,\ldots,n+1$, then $a_1\cdots a_{n+1} \equiv b_1\cdots b_{n+1} \pmod m$. Notice, by assumption, that $a_{n+1} \equiv b_{n+1} \pmod m$. By induction hypothesis, $a_1\cdots a_n \equiv b_1\cdots b_n \pmod m$. Hence, by the above fact, the result follows immediately.

Thus, we conclude, by mathematical induction, that if $a_i \equiv b_i \pmod m,i=1,2,\ldots,k$, then $a_1\cdots a_k \equiv b_1\cdots b_k \pmod m$.

For the second part, let $a \equiv b \pmod m$. Let $k$ be arbitrary positive integer. We'll show that $a^k \equiv b^k \pmod m$. To this end, choose $a_i=a$ and $b_i=b$ for all $i=1,\ldots,k$ and apply to the first part. We obtain that if $a \equiv b \pmod m$, then $a\cdots a (\text{$k$ times}) \equiv b\cdots b(\text{$k$ times})$, which is $a^k \equiv b^k \pmod m$. Hence, proved. Q.E.D.

Does the above approach correct? Any helps would be appreciated. Thanks in advanced.

As long as you can add that proof of $ac\equiv bd$ in the proof, that is literally almost perfect. However, I’ll have to vote to close this question as a duplicate of this as the first (top score) answer contains everything you may need. — insipidintegrator, Sep 17 '22 at 16:37
@insipidintegrator OK, I'll add it now. Please don't go anywhere. — , Sep 17 '22 at 16:39
Does this answer your question? Divisibility by 7 rule, and Congruence Arithmetic Laws (NOTE: This was actually an auto-message posted when I voted to close. I have said the above line in my previous comment too.) — insipidintegrator, Sep 17 '22 at 16:40
@insipidintegrator Unfortunately, it doesn't answer my question yet. — , Sep 17 '22 at 16:44
It proves the “fact” (which, in formal mathematical terms, is called a LEMMA by the way, just some tips) in a slightly different way. — insipidintegrator, Sep 17 '22 at 16:45
@insipidintegrator OK, thanks a lot. But, how about my approach? — , Sep 17 '22 at 16:46
Yes, that's a correct proof of (the inductive extension of) the Congruence Product Rule in the linked dupe. For SV questions you should specify which step of the proof you have doubts about, and why. — Bill Dubuque, Sep 17 '22 at 17:53
As mentioned here in the 2nd linked dupe, it is a special case of the Polynomial Congruence Rule in the first linked dupe. — Bill Dubuque, Sep 17 '22 at 18:36

If $a_i \equiv b_i \pmod m, i=1,2,\ldots,k$, then $a_1\cdots a_k \equiv b_1\cdots b_k \pmod m$.

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