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Prove by induction that $4^n+5^n+6^n$ is divisible by $15$ for all odd $n\in \mathbb{N}$.

My proof : I'm straight going to the induction hypothesis part.

Let for some odd $m(\gt2)\in \mathbb{N}, 15 | 4^m+5^m+6^m$. Then, $$ 4^m+5^m+6^m = 15q$$ for some $q\in \mathbb{Z}$.

Now, $$4^m = 15q-5^m-6^m$$

Since $m$ is odd then the next odd number is $m+2$.

Now, $$ 4^{m+2} + 5^{m+2} + 6^{m+2} \\ = 4^m \cdot 16 + 5^m \cdot 25 + 6^m \cdot 36 \\ = (15q-5^m-6^m) \cdot 16 + 5^m \cdot 25 + 6^m \cdot 36 \\ = 15 \cdot 16q + 9 \cdot 5^m + 20 \cdot 6^m \\ = 15 \cdot 16q + 9 \cdot 5 \cdot 5^{m-1} + 20 \cdot 6 \cdot 6^{m-1} \\ = 15\big( 16q + 3 \cdot 5^{m-1} + 8 \cdot 6^{m-1} \big) \\ \\ \implies 15 | 4^{m+2} + 5^{m+2} + 6^{m+2} $$

Thus from induction the statement is true for every odd $n\in \mathbb{N}$

Is my proof correct?

Bill Dubuque
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Itachi
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    That works. Or, without induction, $,4,5,6,$ are $,1, -1, 0 \pmod 3,$ and $,-1,0,1 \pmod 5,$. – dxiv Feb 12 '22 at 23:17
  • @dxiv Thanks for confirmation. – Itachi Feb 12 '22 at 23:17
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    Stab in the dark: are you concerned about the $m$ to $m + 2$ thing? If it makes you more comfortable, you could try to prove $15 \mid 4^{2n+1} + 5^{2n+1} + 6^{2n+1}$ for $n \ge 0$. That way, this is a traditional induction proof over the natural numbers, and it is appropriate to assume this holds for $n = m$, and prove it holds for $n = m + 1$. Of course, this is still doing basically what you're doing in your proof, just slightly less elegantly. – Theo Bendit Feb 12 '22 at 23:29
  • @Theo actually in my answer I did what you talked about in your comment. Here I had to get the confirmation first so in hurry I did the rough work that came to my mind instantly. – Itachi Feb 12 '22 at 23:38
  • @Theo should I edit or leave it as it is? – Itachi Feb 12 '22 at 23:40
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    @Itachi This answer is fine. Doing it with $2n+1$ is fine too. I personally slightly prefer what you've written, but that's purely a taste thing. Don't worry about editing it. You clearly understand how the induction step works for this problem. – Theo Bendit Feb 12 '22 at 23:48
  • @Theo alright mate! – Itachi Feb 12 '22 at 23:53

3 Answers3

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If you know about linear recurrence: Let $f(n)=4^n+5^n+6^n,$ then you get the recurrence $$\begin{align} f(n+3)&=(4+5+6)f(n+2)-(4\cdot 5+4\cdot 6+5\cdot 6)f(n+1)+4\cdot 5\cdot 6 f(n)\\ &=15f(n+2)-74f(n+1)+240f(n)\\&\equiv f(n+1)\pmod{15} \end{align} $$ So not only is $$f(2k+1)\equiv f(1)=15\equiv 0\pmod {15},$$ but $$f(2k)\equiv f(2)=77\equiv 2\pmod{15},$$ for $k\geq 1.$

If $a,b,c$ are integers, and $g(n)=a4^n+b5^n+c6^n,$ you will get the same congruence, $g(n+3)\equiv g(n+1)\pmod {15}.$

Even more generally, if $a,b,c,d$ are integers and $3\not\mid d,$ and $h(n)=a(d-1)^n+bd^n+c(d+1)^n,$ you also get:

$$h(n+3)\equiv h(n+1)\pmod {3d}$$

Thomas Andrews
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Just a small detail. You proved the induction step $P(m) \implies P(m+2)$ with the hypothesis $m$ odd, $m>2$. If you prove the induction base $m=1$, your proof will be, strictly speaking, incomplete. Because you would have proved

  • $P(1)$
  • $P(m) \implies P(m+2)$ for odd $m > 2$

which translate in proving:

  • $P(1)$
  • $P(3) \implies P(5)$
  • $P(5) \implies P(7)$
  • ...

As you see, you would had missed a link in the chain of implications: $P(1) \implies P(3)$.

You should drop the constraint $m>2$ (that you actually didn't use in your proof).

If you had proved $P(m-2) \implies P(m)$ then the constraint $m > 2$ would be necessary.

jjagmath
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  • Actually I posted the rough work that I came up at first. Then while answering it formally in my notebook I rephrased everything and Proved it in the right way which is : For any odd $n\in \mathbb{N}, \ n = 2m-1, \text{for some} \ m\in \mathbb{N}$. Thus let, $P(m) $ be the statement that $4^n+5^n+6^n$ is divisible by $15$ where $n=2m-1$. Then I followed the induction steps for $m =1, 2, k(\gt 2) \text{and} \ k+1$. – Itachi Feb 12 '22 at 23:52
  • @Itachi Just to be clear. That is not "the" right way, but "a" right way. Proving $P(1)$ and $P(m) \implies P(m+2)$ for $m\ge 1$ is a perfect valid way too. – jjagmath Feb 13 '22 at 00:16
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Alternatively, let $f(n)=4^n+5^n+6^n$. Then $$ f(n+2)-f(n)=15 \cdot 4^n + 24 \cdot 5^n + 35 \cdot 6^n = 15 \cdot 4^n + 15 \cdot 8 \cdot 5^{n-1} + 15 \cdot 14 \cdot 6^{n-1} $$ So if $15$ divides $f(n)$, then $15$ divides $f(n+2)$. Note that $n+2$ is the next odd number if $n$ is odd.

lhf
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