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So the question is as above is shown:

How to prove $8^{2^n} - 5^{2^n}$ is divisible by $13$?

$n=1: 8^2 - 5^2 = 39$ is a multiple of $13$.

I.H.: Suppose $8^{2^n} - 5^{2^n}$ is divisible by $13$ is true $\forall n \in \mathbb{N}\setminus\{0\}$.

I got stuck on this $n= m + 1$ case.

$8^{2^{m+1}} - 5^{2^{m+1}}= 8^{2^m\cdot 2}- 5^{2^m\cdot 2}$

Can somebody please help me?

Martin Sleziak
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Anonymous196
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8 Answers8

5

More generally, $8^{m} - 5^{m}$ is divisible by $13$ when $m$ is even.

Indeed, $8^{m} - 5^{m} = (13-5)^m-5^m = 13a+5^m-5^m=13a$ by the binomial theorem.

Here is a proof by induction:

$8^{m} - 5^{m}=13a$

$8^{2} - 5^{2}=13b$

$8^{m+2} - 5^{m+2} = 8^m 8^2 - 5^m 5^2 = (13a+5^m)8^2 - 5^m 5^2 = 13c + 5^m 8^2 - 5^m 5^2 = 13c +5^m(8^2 - 5^2) = 13c + 5^m 13b = 13d$

lhf
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For $n>1$ $${(8^2)}^{2^{n-1}}-{(5^2)}^{2^{n-1}}=64^{2^{n-1}}-25^{2^{n-1}}\equiv(-1)^{2^{n-1}}-(-1)^{2^{n-1}}\equiv0~~~\text{mod}~13$$ case $n=1$ is trivial.

Nosrati
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Hint: $8^{x \cdot 2}-5^{x \cdot 2}=(8^x)^2-(5^x)^2=(8^x-5^x)(8^x+5^x)$

Fred
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$\bmod 13\!:\,\ 8\equiv -5\,\overset{(\ \ )^{\Large 2}}\Longrightarrow\, 8^{\large 2}\!\equiv 5^{\large 2}\overset{(\ \ )^{\Large N}\!}\Longrightarrow\, 8^{\large 2N}\!\equiv 5^{\large 2N}\,$ by the $ $ Congruence Power Rule.

Remark $ $ Replying to comments, below we show how the inductive proof of the linked Power Rule can be expressed without knowledge of congruences, using analogous divisibility rules.

$\begin{align}{\bf Divisibility\ Product\ Rule}\ \ \ \ &m\mid \ a\ -\ b\qquad {\rm i.e.}\quad \ \, a\:\equiv\: b\\ &m\mid \ \ A\: -\: B\qquad\qquad \ A\equiv\, B\\ \Rightarrow\ \ &\color{}{m\mid aA - bB}\quad \Rightarrow\quad aA\equiv bB\!\pmod{\!m}\\[.2em] {\bf Proof}\,\ \ m\mid (\color{#0a0}{a\!-\!b})A + b(\color{#0a0}{A\!-\!B}) &\,=\, aA-bB\ \ \text{by $\,m\,$ divides $\rm\color{#0a0}{green}$ terms by hypothesis.}\end{align}$

$\begin{align}{\bf Divisibility\ Power\ Rule}\qquad &m\mid a\ -\ b\qquad {\rm i.e.}\qquad a\equiv b\\ \Rightarrow\ \ & m\mid a^n-b^n\quad\ \Rightarrow\quad\,\ \ a^n\!\equiv b^n\pmod{\!m} \end{align}$

Proof $\ $ The base case $\,n=0\,$ is $\,m\mid 1-1\,$ so true, and the inductive step follows by applying the Product Rule, namely $\ m\mid a- b,\,a^n- b^n \Rightarrow\, m\mid a^{n+1}-b^{n+1}.\,$

Your exercise follows from the specialization $\,a = 8^2,\ b = 5^2,\ m = 13\, $ in the above Power Rule. More explicitly, note that we have $\,\color{#c00}{13}\mid 8^2-5^2 = (\color{#c00}{8\!+\!5})(8\!-\!5),\,$ so powering that yields

$\begin{align}{\bf Divisibility\ Power\ Rule}_{\,8,5}\quad &13\mid 8^2 - 5^2\qquad {\rm i.e.}\qquad 8^2\equiv\, 5^2\\ \Rightarrow\ \ & 13\mid 8^{\large 2n}\!-\!5^{\large2n}\quad\ \Rightarrow\quad\,\ \ 8^{\large 2n}\!\!\equiv 5^{\large 2n}\!\!\!\pmod{\!13} \end{align}$

Thus it holds true for all even exponents $2n,\,$ which includes your expoents $2^n,\ n\ge 1.$

Bill Dubuque
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  • The Power Rule has a simple inductive proof - see the lined post. – Bill Dubuque Sep 11 '17 at 14:12
  • Yeah I've already remarked that with modular arithmetic. I wanted it proof the above expression without using this method. Thanks for the tip though. I should have said that earlier – Anonymous196 Sep 11 '17 at 14:15
  • @AnonymousI Then you can use $, a-b\mid a^n-b^n,$ for $,a=8^2,, b = 5^2,,$ which is a special case of the Polynomial Factor Theorem $, x-y\mid f(x)-f(y),,$ which has a very simple inductive proof using the Polynomial Division Algorithm (follow the link). – Bill Dubuque Sep 11 '17 at 14:23
  • With my previous remark I meant that I only came up with 8 is congruent with -5( mod 13) and I knew that if you took the power of the congruent those were the same mod 13. And then I stopped without realizing what you wrote at last. It was helpful. – Anonymous196 Sep 11 '17 at 14:34
  • @AnonymousI Generally $\ a\equiv b,\Rightarrow, f(a)\equiv f(b),$ for any polynomial $,f(x),$ with integer coefficients (above is special case $,f(x) =x^n).,$ This Polynomial Congruence Rule. is the congruence analog of the Polynomial Factor Theorem. While you can prove special cases of these results without congruences, they ususally end up being obfuscated unwindings of the simple congruence-based proofs. I strongly recommend mastering the congruence methods, since without them you will soon be lost in more complicated cases. – Bill Dubuque Sep 11 '17 at 14:40
  • I will when I get to that chapter in my book. I'm still at chapter 1, and the congruence methods is at chapter 4. – Anonymous196 Sep 11 '17 at 14:43
  • @AnonymousI Before congruences are known you can still take advantage of these ideas by using the Congruence Product Rule in divisibility form as I explain here. Inductively iterating the Product Rule yields the Power Rule. This is slightly more cumbersome than using congruences, but usually much more insightful than a brute-force induction proof (which usually greatly obscures the arithmetical essence of the matter - which here is simply that congruenes (equations) remain true when raised to integer powers). – Bill Dubuque Sep 11 '17 at 14:51
  • So I've read al of your referred posts which are really clear. So to ask you more on that way of proving. Does it suffice to say that $8^{2^n}≡5^{2^n} \pmod {13}$? Forgive me for my bad MathJax. – Anonymous196 Sep 11 '17 at 15:21
  • @AnonymousI All of the methods I mention allow us raise to power $N$ the congruence $,8^2\equiv 5^2,$ to deduce $,8^{2N}!\equiv 5^{2N}!\pmod{!13},,$ or, in divisibility language $,13\mid 8^2-5^2,\Rightarrow, 13\mid 8^{2N}-5^{2N}.,$ Yours is a special case since it has even power $, 2^n = 2N$ by $,n > 0\ \ $ – Bill Dubuque Sep 11 '17 at 15:24
  • Let us continue this discussion in chat. – Anonymous196 Sep 11 '17 at 15:30
  • @AnonymousI If there is anything that remains unclear then let me know and I'll be happy to elaborate further. – Bill Dubuque Sep 13 '17 at 19:12
  • It looks like you're reading my mind. – Anonymous196 Sep 13 '17 at 19:56
  • I'm just stuck with the finish of the proof – Anonymous196 Sep 13 '17 at 19:57
  • Does the end of the proof look like this: $8^{2^{n+1}} \equiv -5^{2^{n+1}} \pmod {13}$? – Anonymous196 Sep 13 '17 at 19:59
  • And the other alternative proposed by user Ihf. Instead of using $m$ I used $m+1$ as exponent for the 1st equation and $3$ as the exponent for the 2nd equation. But I got stuck at the end of the equation. I'll show you my proof. – Anonymous196 Sep 13 '17 at 20:06
  • @AnonymousI I edited the answer to show more explicitly how to conclude the proof by the Power Rule. Note that there is no explicit induction when proved this way (it is encapsulated in the inductive proof of the Power Rule). The 2nd proof in lhf's answer essentially repeats the the inductive proof of the Power Rule in this special case (but that way makes it much harder to see the arithmetical essence of the matter, i.e. that it boils down to simply powering both sides of an equation = congruence). – Bill Dubuque Sep 13 '17 at 22:17
  • @AnonymousI Hopefully it is now cleaer. Congruences are generalizations of equalities. You might find it helpful to consider the equality analog of the above, namely $, a = -b,\Rightarrow, a^2 = b^2,\Rightarrow, a^{2n} = b^{2n},$ by induction on $n.,$ The above is the same except we replaced equality by congruence, i.e. we generalized it from integers to integers $!\bmod m$ (the integer case is a special case of the modular generalization by choosing the modulus $,m=0,$ since $,a\equiv b\pmod{0}\iff 0\mid a-b \iff a-b=0\iff a=b,,$ so the integers $!\bmod 0$ is just the integers – Bill Dubuque Sep 14 '17 at 04:06
  • Please see my continuation of Ihf's suggestion in my answer below. As for your suggestion I'll post it. – Anonymous196 Sep 14 '17 at 11:16
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You could prove by induction that $f(n)=8^{2^n}\equiv1\bmod{13}$ holds for $n\ge3$. Indeed, for $n=2$ we have $f(2)=64\equiv-1\bmod{13}$. Since $f(n+1)\equiv f^2(n)\bmod{13}$, the claim follows.

Analogously, you can prove that $g(n)=5^{2^n}\equiv1\bmod{13}$ for $n\ge3$.

Combining these two observations proves your statement for $n\ge3$. You may easily check the remaining two cases $n=1$ and $n=2$ by hand.

Gamow
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To complete the proof with the following identity $ A^2 - B^2 = (A + B) (A-B) $. I get $\begin{align}\\ \underline {\text{n= m+1:}} \\ 8^{2^n} -5^{2^n} & = 8^{2^{m+1}} -5^{2^{m+1}} \\ & = (8^{2^m} + 5^{2^m})(8^{2^m} -5^{2^m})\\ & = (8^{2^m} + 5^{2^m})\cdot \text{some multiple of 13} \end{align}$

Thus concluding that the entire expression is divisible by 13.

Anonymous196
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    You have got the incorrect $A$ and $B$ when using $A^2 - B^2 = (A-B)(A+B)$. –  Sep 15 '17 at 10:08
  • Stupid of me I've got it on my paper but I didn't edit it there. – Anonymous196 Sep 15 '17 at 13:15
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    I mean this one: $8^{2^{m+1}} -5^{2^{m+1}} = (8^{m+1} + 5^{m+1})(8^{m+1} -5^{m+1})$. It's incorrect. Note $8^{2^{m+1}} - 5^{2^{m+1}} = (8^{2^m})^2 - (5^{2^m})^2$. –  Sep 15 '17 at 14:27
  • Oh, ok. I see now – Anonymous196 Sep 15 '17 at 19:03
  • I will change it – Anonymous196 Sep 15 '17 at 19:14
  • "For the expression in the first paranthesis I've noticed that it's divisible by 13 if and only if $m$ is an even number" No..... Forget all the other answers and go back to your original question. Use the I. H. –  Sep 15 '17 at 21:33
  • Yeah because the exponents will be of the power series of 2. So both parantheses hold for $m\in \mathbb N_0$ – Anonymous196 Sep 15 '17 at 21:40
  • So if I substitute the paranthesis $8^{2^m} -5^{2^m}$ with the I.H. I conclude that the other paranthesis is a multiple of 13. And then I'm done? – Anonymous196 Sep 15 '17 at 21:43
  • @JohnMa Is my edited proof now sufficient to be a good inductive proof. I do think it needs some better formal notes but I'm quite pleased with the end of it. – Anonymous196 Sep 15 '17 at 22:46
  • Yes, perfect, you are done. Next time if you want someone to critique your work, do not write an answer, instead edit your question and adds a comments to the answers to which you want the answerers to respond. Or they will never see your answer. –  Sep 15 '17 at 23:09
  • I will do that in the near future. Thanks for your confirmation. – Anonymous196 Sep 15 '17 at 23:24
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Because $$8^{2^n}-5^{2^n}=64^{2^{n-1}}-25^{2^{n-1}}=(64-25)\left(64^{2^{n-1}-1}+...+25^{2^{n-1}-1}\right)$$ is divisible by $39$ and $39$ is divisible by $13$.

Michael Rozenberg
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Consider the product $$(8-5)(8+5)(8^2+5^2)(8^{2^2}+5^{2^2}) \dots(8^{2^{n-1}}+5^{2^{n-1}})$$which is a form of "telescoping" product. The second factor is equal to $13$. This is equivalent to some other observations people have made, of course, but can be a useful way of looking at things from a different angle.

Mark Bennet
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  • Yeah I've come across telescoping sum and products before but I'd never think of it to use it here in the proof. Interesting way to see that. But what do you need to do to complete this proof? Does one just say because 13 is a product in this telescoping product that is indeed divisible by 13 and then the proof is finished? – Anonymous196 Sep 11 '17 at 15:27
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    See this answer for a vivid view of the telescopic cancellation in the above answer. – Bill Dubuque Sep 12 '17 at 15:02
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    @AnonymousI Good question. To be rigorous we'd need to prove by induction that $,13\mid 13, a_1\cdots a_n,,$ e.g. by proving that the product of $n$ integers is an integer (possibly a corollary of an earlier (rigorous) recursive definition of products $,\prod a_i)$. It's instructive to compare to the congruence method: it works simply by squaring $,8^{\large 2^n}!\equiv 5^{\large 2^n}$ to get the $n+1$ case, but the above computes their quotient and accumulates it into the total quotient. But computing the quotient is usually (much) wasted effort if only the remainder is needed. – Bill Dubuque Sep 12 '17 at 15:07