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The congruence equation $a\equiv b \pmod n \implies \exists k \in \mathbb{Z}, a -b = kn.$
Taking the i-th power of both sides : $(a -b)^i = k^in^i$.


$(a-b)(a-b)^{i-1}=k^in^i$.
$n\mid (a-b), (a-b) \mid (a-b)^i \implies n\mid (a-b)^i \implies \exists m \in \mathbb {Z}, (a-b)^i = mn$.

But, my proof is incomplete, as it does not show still that $a^i - b^i = kn$.

All the comments and answer have ignored the fact that how to get the term $a^i - b^i$ in the first place.
It seems that all have taken the approach to simply take the $a^i- b^i=kn$ as a new equality, not derived from the original one. Just it uses the property of $n \mid (a-b)$ of the original one.

jitender
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    How on Earth have you switched from $a^i-b^i$ to $(a-b)^i$ ... and back ... ? –  Feb 10 '18 at 23:30
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    Use the fact that $a^i-b^i=(a-b)(a^{i-1}+a^{i-2}b+a^{i-3}b^2+...+a^{2}b^{i-3}+ab^{i-2}+b^{i-1})$ – rtybase Feb 10 '18 at 23:30
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    This is unreadable. Please edit. – lulu Feb 10 '18 at 23:33
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    If you know that $(a-b)|(a^i-b^i)$ and you know that $n|(a-b),$ aren't you done? What is all this other stuff? – saulspatz Feb 10 '18 at 23:35
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    Your proof makes no sense. Either use the binomial theorem or use a simple induction. – lulu Feb 10 '18 at 23:36
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    Are you asking a question? Are your writing a proof? What IS this post? Why did you post this? – fleablood Feb 10 '18 at 23:43
  • @fleablood Wanted to confirm the proof, as was not clear about. I was confused as $n\mid (a-b), n\mid n\n^i$ did not seem to imply. I needed transtivity property of division, as shown by 'saulspatz'. – jitender Feb 10 '18 at 23:43
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    @jitender, amWhy deleted his comment ... because 6^3=216 – rtybase Feb 10 '18 at 23:46
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    As a proof, it is completely unreadable. And is completely wrong. $(a-b)^i \ne a^i - b^i$. "(a-b) after taking out other terms = $k^in^i$" doesn't make any sense; it doesn't even seem to be an english sentence. And why would final line $n|a-b, n|n^i\imply (a^i - b^i)$. – fleablood Feb 10 '18 at 23:48
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    @jitender Do you know the binomial theorem? Have you seen my answer? The equality $a^i - b^i = pn$ came from $a - b = kn$ – bames Feb 11 '18 at 00:26
  • $n|(b-a)$ that was given. So $b-a = nk$ for some $k$. $(b-a)|(b^i - a^i)$; you seem to accept that. So $b^i - a^i = (b-a)j$ for some $j$... in fact $j = (b^{i-1} + b^{i-2}a + ....+ba^{i-2} + a^{i-1})$. Soooooo..... $b^i - a^i = (b-a)j = (nk)j = n(kj)$. Therefore $n|(b^i - a^i)$ That's it. You are done. You have shown everything*. – fleablood Feb 11 '18 at 00:31
  • @bames Thanks a lot. I read it earlier, but could not understand at that time the crux. I am very indebted for making the answer and clarifying its contents now. I hope my edited OP has no logical errors otherwise. Please vet it, as seem to err a lot in its framing earlier. ---I mean that is taking the power $(a-b)^i = k^in^i$ logically wrong. – jitender Feb 11 '18 at 00:32
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    @fleablood I think he wanted to prove this purely by modifying $a-b = kn$. It is not the prettiest way to prove it, I agree, but that is what he seems to want. – bames Feb 11 '18 at 00:43
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    @jitender It is true that if $a-b = kn$, then $(a-b)^i = k^in^i$. So that is fine. – bames Feb 11 '18 at 00:46
  • @bames Although, unrelated but have a secondary question. As, both $a^i- b^i = kn, (a-b)^i = k^in^i$ are true, so there should be some sort of equivalence between $a^i - b^i$ and $(a-b)^i$. – jitender Feb 11 '18 at 00:53
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    @jitender they are both in the equivalence class $\bar 0$ of $\Bbb Z_n$ – bames Feb 11 '18 at 00:57
  • @bames So, I hope you mean: (i)By $(a-b)^i = k^in^i$, can derive : $n \mid n^i, n^i \mid (a-b)^i \implies n \mid (a-b)^i$. (ii) $n \mid (a^i-b^i)$. By (i) , (ii) both $(a^i - b^i) $, & $(a-b)^i$ are in the same residue class w.r.t. $n$. But, it is not complete, & if need a rigorous proof, then how to proceed. – jitender Feb 11 '18 at 01:05
  • Let us continue this discussion in chat. – bames Feb 11 '18 at 01:06
  • Right. $b-a=kn $ so $b^i-a^i=kn*(b^{i-1}+ab^{i-2}+...+a^{i-1} )$. So what on earth is the issue? – fleablood Feb 11 '18 at 05:04

5 Answers5

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This is not the easiest answer but hope it will be instructive, as it illustrates some important points:

Lemma: If $a\equiv b\pmod n$ and $c\equiv d\pmod n$, then $ac\equiv bd\pmod n$. (Congruences can be multiplied.)

Proof of the Lemma: $bd-ac=bd-ad+ad-ac=(b-a)d+a(d-c)$. If $n\mid b-a$ and $n\mid d-c$, i.e. $b-a=un, d-c=vn$, then $bd-ac=und+avn=(du+av)n$ and so $n\mid bd-ac$, i.e. $ac\equiv bd\pmod n$.

Proof of the claim: by induction on $i$:

  • $i=0$: $a^0=1\equiv 1=b^0\pmod n$ - trivially
  • $i\to i+1$: If $a\equiv b\pmod n$ and $a^i\equiv b^i\pmod n$ (inductive hypothesis), then by using Lemma, and multiplying those congruences, we get the congruence $a^{i+1}\equiv b^{i+1}\pmod n$.
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It might be easier to use $a = kn + b$ and the Binomial theorem. Then $a^i = (kn+b)^i = \sum_{j=0}^i \binom{i}{j}(kn)^jb^{i-j}$. All of these terms have a factor of $n$ in them except for the $0^{th}$ one, so this equation may be written in the form $a^i = pn + b^i$. Hence the claim.

bames
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I think it is easiest first to prove

Prop: If $a \equiv b \mod n$ and $c \equiv d \mod n$ then $ac \equiv bd \mod n$.

Proof: $n|(a-b)$ so $a-b = kn$ so $a = b + kn$ for some integer $k$. And $n|c-d$ so $c-d = jk$ so $c = d + jn$ for some integer $j$. So $ac =(b+kn)(d+jn) = bd + kdn + bjn + kjn^2 = bd + n[dk + bj + kjn]$ so $ac - bd = n[dk + bj + kjn]$ so $n|ac-bd$ so $ac\equiv bd \mod n$.

Then by induction, If $a_i \equiv b_i \mod n$ for several $a_i$ and $b_i$ then $a_1*a_2*..... * a_n\equiv b_1*b_2*....*b_n \mod n$.

SO if $a\equiv b \mod n$ then $a*a \equiv b*b \mod n$ and $(a*a)*a \equiv (b*b)*b \mod n$, etcc. so $a^i \equiv b^i \mod n$.

==== or =====

$a\equiv b \mod n$ means $(b-a) = n*k$ for some integer $k$.

$b^i - a^i = (b-a)(b^{i-1}+ b^{i-2}a + ..... + ba^{i-2} + a^{i-1})$

So $b^i - a^i = (n*k)(b^{i-1}+ b^{i-2}a + ..... + ba^{i-2} + a^{i-1})$

$= n*[k(b^{i-1}+ b^{i-2}a + ..... + ba^{i-2} + a^{i-1})]$.

Now, clearly you must agree than $[k(b^{i-1}+ b^{i-2}a + ..... + ba^{i-2} + a^{i-1})]$ is an integer!

SO $n|(b^i - a^i)$.

So $a^i \equiv b^i \mod n$.

fleablood
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$$a\equiv b \pmod n \implies n|(b-a)$$

$$n|(b-a),\text { and } (b-a)|(b^i-a^i) \implies n| (b^i-a^i) $$

$$ n| (b^i-a^i) \implies a^i\equiv b^i \pmod n$$

Mohammad Riazi-Kermani
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    How do we know $(b-a)|(b^i - a^i)$. I don't think that is obvious to a student at the level of the OP. – fleablood Feb 10 '18 at 23:48
  • @fleablood It is clear that the polynomial $(b^i-a^i)$ has factorization as $(b-a)(b^{i-1} + b^{i-2}a + b^{i-3}a^2+\ldots +a^{i-1})$. What is not clear is how to get $b^i - a^i$ out of the original expression. I mean, as also shown in the OP, there need to be taken exponents of both sides $(a-b) = kn$, so how $a^i - b^i$ results. – jitender Feb 11 '18 at 00:16
  • If you know that $(b-a)|(b^i - a^i)$ then you are done! $n |(b-a)$ and $(b-a)|(b^i - a^i)$ then $n|b^i - a^i$! So, I don't know what you mean "get $b^i - a^i$ out of the original expression". That isn't an issue if you know $b-a|b^i - a^i$ – fleablood Feb 11 '18 at 00:27
  • @fleablood I wish and request you to please elaborate why it is not an issue to know how $b^i - a^i$ is generated from the original expression. Your comment is not clear about that, and seemingly emphasizes end result. – jitender Feb 11 '18 at 02:22
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    I can't explain why something ISN'T an issue. It's your job to explain why you think it is. If $a|b $ and $b|c $ then $a|c $. So if $n|b-a $ and $b-a|b^i -a^i $, then $n|b^i-a^i $. Why isn't that good enough for you? – fleablood Feb 11 '18 at 05:01
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$a≡ b\mod n ⇒ a= k.n +b; k∈ N $

⇒ $a^i=(kn+b)^i= k_1.n+b^i; k_1∈ N ⇒ a^i ≡ b^i \mod n$

sirous
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