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I am trying to prove by induction that $\sum_{i=3}^{5} i^n$ is divisible by $12$ for all positive odd integers $n$. I heard that the solution might involve normal induction, where you assume that it works for the lowest positive odd number (in this case, $1$) and then try to deduce $P(n+2)$ for the next odd number.

I am not sure how to go about this, so any help would be greatly appreciated. I was thinking that if we define $n$ to be odd, then we could try to show that $n+2$ works using the fact that $n+2$ is also odd. However, I am not sure if this is the correct approach or how to proceed.

Could someone please help me with this problem?

Thank you!

VTand
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Bryan C
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    You can try setting $n=2k+1$ for $k\in\mathbb{Z}$ and do the induction on $k$ instead. Now think about how to show whether something is divisible by $12$. – Landuros Dec 16 '22 at 03:44
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    You could try looking at the powers of $3,4$ and $5$ $\mod 12$ and how they add up. – Jonah Dec 16 '22 at 03:48
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    If you are allowed to use strong induction, use the identity $$x^{n+3}+y^{n+3}+z^{n+3}=(x+y+z)(x^{n+2}+y^{n+2}+z^{n+2})-(xy+yz+zx)(x^{n+1}+y^{n+1}+z^{n+1})+xyz(x^n+y^n+z^n).$$ – Bumblebee Dec 16 '22 at 04:16
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    $\color{#c00}{\text{1-liner:}}$ $!\bmod 3 ,&, 4$ it is $\equiv 1^n + (-1)^n \equiv 0,$ so it's divisible by $3$ & $4$ hence also by their lcm = $12$. $\bf\ QED$ This is a proof by induction on $n$ (it's hidden in a full proof of the claim that $(-1)^n\equiv -1$ for odd $n,,$ by the Congruence Power Rule, which is proved by induction on $n).,$ See here and here for further detail. – Bill Dubuque Dec 16 '22 at 09:24
  • $x_n=3^n + 4^n + 5^n$ satisfies $x_{n+3}=12 x_{n+2} - 47 x_{n+1} + 60 x_n$. The equation comes from expanding $(x-3)(x-4)(x-5)$. – lhf Dec 16 '22 at 10:33
  • Alternatively, compute $(3^m + 4^m + 5^m)-(3^n + 4^n + 5^n)$ for $m=n+2$ – lhf Dec 18 '22 at 21:58

4 Answers4

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We set $n = 2k - 1$ and carry an induction over $k$. For the base case, $k = 1$, $n = 1$ and $3 + 4 + 5 = 12$, which is clearly divisible by 12. Assume true for $k$, and consider the case for $k + 1$. We have. \begin{align*} 3^{2k + 1} + 4^{2k + 1} + 5^{2k + 1} & = 9\cdot 3^{2k-1} + 16\cdot 4^{2k-1} + 25\cdot 5^{2k-1}\\ & = 8\cdot 3^{2k-1} + 15\cdot 4^{2k-1} + 24\cdot 5^{2k-1} + (3^{2k-1} + 4^{2k-1} + 5^{2k-1}). \end{align*} From our induction hypothesis, we know $3^{2k-1} + 4^{2k-1} + 5^{2k-1} \equiv 0 \ \ \ (\text{ mod } 12 \ )$. Also, $8\cdot 3^{2k-1}$, $15\cdot 4^{2k-1}$ and $24\cdot 5^{2k-1}$ are all $\equiv 0 \ \ \ (\text{ mod } 12 \ )$ implying $3^{2k + 1} + 4^{2k + 1} + 5^{2k + 1}$ is divisible by 12.

sudeep5221
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  • You said consider the case for k+1, but you wrote 2k+1, why? And why do we set n = 2k-1? – Bryan C Dec 16 '22 at 06:50
  • This is because we are only considering odd integers. Setting $n = 2k - 1$ ensures $n$ is odd. For the inductive step, $2k + 1 = n + 2$, like you suggested. – sudeep5221 Dec 16 '22 at 14:05
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Here is a stronger theorem which is easier to prove.

Suppose that we have $a + b + c \equiv 0 \mod 12$, with the criterion that $a$ is divisible by 3 and $b$ is divisible by 4. This is true for $a=3$, $b=4$, and $c=5$, for instance, but also many other things.

We then want to show that $9a + 16b + 25c \equiv 0 \mod 12$ as well. If $a, b,$ and $c$ are odd powers $3, 4,$ and $5$, then these would be the next odd powers, but we've left it open for $a, b, $ and $c$ to be anything such that they sum to $0 \mod 12$ and with $a$ divisible by $3$ and $b$ divisible by $4$.

Then via elementary algebra $\mod 12$, we can show the following:

$$ 9a + 16b + 25c \mod 12 \\ = 9a + 4b + c \mod 12 \\ $$

Now, since $a + b + c \equiv 0 \mod 12$, we can subtract $a + b + c$ from the above to get

$$ 8a + 3b \mod 12 $$

But now we are done, because we know that $a$ is divisible by $3$ and $b$ is divisible by $4$. If we make the variable substitution that $a = 3a'$ and $b = 4b'$, we get

$$ 8\cdot 3a' + 3 \cdot 4b' \mod 12 \\ = 24 a' + 12 b' \mod 12 \\ = 0 a' + 0 b' \mod 12 \\ = 0 \mod 12 $$

So we have proven that for any $a, b, c$ which sum to $0$ mod 12 and for which $a$ and $b$ are divisible by 3 and 4, respectively, that we also have $9a + 16b + 25c \equiv \mod 12$.

Since $a, b, c = 3, 4, 5$ is the base case which fits these criteria, and we know that $9a + 16b + 25c$ also fits these criteria - which are also the next odd powers of $3, 4$ and $5$, we thus have that every sum of odd powers of $3, 4$ and $5$ fits these criteria.

But we also have that this is true for other triples of numbers satisfying the original properties. For instance, we have $a = 15, b = 8, c = 1$ sum to $0 \mod 12$, and so do $9\cdot 15, 16 \cdot 8, $ and $25 \cdot 1$. Anyway, that proves the theorem.

Mike Battaglia
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The base case is easy enough, so we can try to rewrite $P(n+2)$ as you suggest. Assuming $P(n)$ we have

$$3^{n+2}+4^{n+2}+5^{n+2} = 25(3^n +4^n +5^n) - 12(4\cdot 3^{n-1} + 3\cdot 4^{n-1})$$

By the inductive hypothesis the term on the left is a multiple of $12$ so the entire expression must be a multiple of $12$.

Ninad Munshi
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  • I'd probably add explicitly that 3^n + 4^n + 5^n is divisible 12, and I think you have 1 too many 5^n on the right side. Otherwise, looks good. – Bryan C Dec 16 '22 at 04:18
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    I appreciate the comments, but (a) that's what the phrase "assume $P(n)$" means - in the nomenclature you have to familiarize yourself with, $P(n)$ is a logical statement, in this case, our inductive hypothesis that "$3^n + 4^n + 5^n$ is divisible by 12". So the statement "assume P(n)" is a standard shorthand for "assume the inductive hypothesis $3^n + 4^n + 5^n$ is divisible by 12" and (b) $5\times 5 = 25$ – Ninad Munshi Dec 16 '22 at 13:47
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Proof 1 with Induction :

We know $X=3^{1}+4^{1}+5^{1}=12$ is Divisible by $12=3 \times 4$.

Let $Y=3^{n}+4^{n}+5^{n}$ be Divisible by $12$.

Let $Z=3^{n+2}+4^{n+2}+5^{n+2}$ , where we want to check Divisibility by $12$ , starting with $3$.
$Z=3^{2} \times 3^{n}+4^{2} \times 4^{n}+5^{2} \times 5^{n}$
$Z=9 \times 3^{n}+16 \times 4^{n}+25 \times 5^{n}$
$Z=(8+1) \times 3^{n}+(15+1) \times 4^{n}+(24+1) \times 5^{n}$
$Z=(8+1) \times 3^{n}+(15+1) \times 4^{n}+(24+1) \times 5^{n} = 3^{n}+4^{n}+5^{n}+3c=Y+3c$ , which must be Divisible by $3$

Continuing with $Z=3^{n+2}+4^{n+2}+5^{n+2}$ , where we want to check Divisibility by $12$ , which contains $4$.
$Z=3^{2} \times 3^{n}+4^{2} \times 4^{n}+5^{2} \times 5^{n}$
$Z=9 \times 3^{n}+16 \times 4^{n}+25 \times 5^{n}$
$Z=(8+1) \times 3^{n}+(15+1) \times 4^{n}+(24+1) \times 5^{n}$
$Z=(8+1) \times 3^{n}+(15+1) \times 4^{n}+(24+1) \times 5^{n} = 3^{n}+4^{n}+5^{n}+4k=Y+4k$ , which must be Divisible by $4$

Hence overall, $Z$ must be Divisible by $3 \times 4 = 12$.

With $X$ , $Y$ & $Z$ being Divisible by $12$ , we can see that the Induction Proof works here.

More generally , we can see that $(W-1,W,W+1)$ with $W$ a multiple of $4$ will have that Property.
With $W=4$ we get $(3,4,5)$ here.
With $W=8$ we get $(7,8,9)$ which will have that Property.

Proof 2 without Induction :

$(W-1)^n+(W)^n+(W+1)^n$ has 3 consecutive numbers where we must have a multiple of $3$ , the others must have "1 less than a multiple of $3$" & "1 more than a multiple of $3$".
Evaluating each Power will give terms with multiples of $3$ except the last terms which will be $(-1)^{ODD}$ & $(+1)^{ODD}$ which will give $-1+1=0$ hence the whole must be Divisible by $3$.

$(W-1)^n+(W)^n+(W+1)^n$ with $W=4w$ gives :
$(4w-1)^n+(4w)^n+(4w+1)^n$ with $W=4w$
When we evaluate each Power , we will get multiples of $4$ throughout , except for the last terms which will be $(-1)^{ODD}$ & $(+1)^{ODD}$ which will give $-1+1=0$ , hence the whole will be Divisible by 4.

In general , the whole must be Divisible by $3 \times 4 = 12$
In Particular , $W=4$ gives $3^n+4^4+5^n$ Divisible by $12$ when $n$ is ODD.

Prem
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