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Prove by induction:

$4^n+5^n+6^n$ is divisible by 15 for positive odd integers

For $n=2k-1,n≥1$ (odd integer)

$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$

To prove $n=2k+1$, (consecutive odd integer)

$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,

How do I substitute the statement where $n=2k-1$ to the above, to factor out 15 in order to prove divisibility? Would it be easier to assume $n=k$ is odd and prove $n=k+2$ is divisible by 15?

Bill Dubuque
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Andrei Lenedin
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6 Answers6

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Hint

Like Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$

If $f(m)=4^{2m+1}+5^{2m+1}+6^{2m+1},$

$$f(n+1)-4^2f(n)=5^{2n+1}(5^2-4^2)+6^{2n+1}(6^2-4^2)$$ will be clearly divisible by $15$ if $n\ge0$

So, if $15$ divides $f(n),15$ will divide $f(n+1)$

Now establish the base case i.e., $m=0$

lab bhattacharjee
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You could go mod $3$ and mod $5$ and conclude, an alternate proof is by induction : of course $4^n + 5^n + 6^n$ is divisible by $15$ when $n=1$. However, note that if $n \geq 3$: $$ 4^n + 5^n + 6^n - 4^{n-2} - 5^{n-2} - 6^{n-2} \\=\color{blue}{(4^n - 4^{n-2})} + \color{green}{(5^n - 5^{n-2})} + \color{red}{(6^n - 6^{n-2})} \\= \color{blue}{15(4^{n-2})} +\color{green}{ 15(8 \times 5^{n-3})} + \color{red}{ 15(14 \times 6^{n-3})} $$

where terms of same colour are equal by factorization. Thus, the claim follows since the sum of two multiples of $15$ is also a multiple of $15$.

Sarvesh Ravichandran Iyer
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Hint for proof by induction:

$4^{2k+1}+5^{2k+1}+6^{2k+1}=16(4^{2k-1}+5^{2k-1}+6^{2k-1})+9\times5^{2k-1}+20\times6^{2k-1}$

As I said in a comment to the question, it's easy to prove with modular arithmetic,

because mod $3$ it's $1+(-1)+0$, and mod $5$ it's $(-1)+0+1$.

J. W. Tanner
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As you suggested, it's notationally simpler to suppose $4^k+5^k+6^k$ is divisible by $15$ and consider $$4^{k+2}+5^{k+2}+6^{k+2} = 16\cdot 4^k + 25\cdot 5^k + 36\cdot 6^k.$$ Subtracting the original expression, we get $15\cdot 4^k + 24\cdot 5^k + 35\cdot 6^k$. The first term is divisible by $15$. Now note that $$24\cdot 5^k +35\cdot 6^k = 15\cdot 8\cdot 5^{k-1} + 15\cdot 14\cdot 6^{k-1}$$ is likewise divisible by $15$. Thus, $4^{k+2}+5^{k+2}+6^{k+2}$ is indeed divisible by $15$.

Query: Where did we use that $k$ is odd? Well, obviously to start the induction. But where else?

Ted Shifrin
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  • "Where did we use that k is odd? Well, obviously to start the induction. But where else?" There doesn't need to be any place else. The induction step was an increase of $2$ so it will only "hit" odd numbers. In fact if you did a base case of $k=2$ and did and induction step of $2$, you will prove that if $n$ is even it won't be divisible by $15$ (because the difference is and the base case, $n=2$, isnt. – fleablood Nov 21 '19 at 14:25
  • An increase of $1$ not two would have $4^{k+1}+5^{k+1}+6^{k+1}=44^k + 55^{k} + 66^k$ and subtracting the original get $45^k + 56^k$ which certainly is not* divisible by $15$ as both terms are divisible by $5$ but one and not the other is by $3$. SO you don't use that $k$ is odd, but you do prove it is only true if $k$ is odd. – fleablood Nov 21 '19 at 14:32
  • @fleablood: You're right. I was thinking that I need to know that $k-1\ge 0$, but if $k$ were a positive even integer, that would still be true. My fault. – Ted Shifrin Nov 21 '19 at 16:21
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$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$

How do I substitute the statement where n=2k−1 to the above

By factoring one more power out...

$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}=(16)4^{2k-1} + (25)5^{2k-1} + (36)5^{2k-1}$

So this is $[16(4^{2k-1} + 5^{2k-1}+6^{2k-1})] + 9*5^{2k-1} + 20*6^{2k-1}$.

And it's easy to finish:

$=[16*15N] + 3*15*5^{2k-2} + 4*15*2*6^{2k-2}$.

====

But if you know modulo arithmetic this is CUTE!

$4^{n} + 5^n + 6^n = (3+1)^n + (6-1)^n + 6^n \equiv 1^n+(-1)^n + 0^n \equiv 0 \pmod 3$ so $3|4^n + 5^n +6^n$.

And $4^n + 5^n + 6^n = (5-1)^n + 5^n + (5+1)^n\equiv (-1)^n + 0^n + 1^n \equiv 0 \pmod 5$ so $5|4^n + 5^n +6^n$.

So $15|4^n + 5^n +6^n$.

....

If you don't know modulo arithmetic the you can use binomial theorem.

$4^n + 5^n + 6^n =(5-1)^n + 5^n + (5+1)^n =$

$(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... +n*5 - 1) + 5^n +(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... -n*5 + 1)=$

$(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... +n*5) + 5^n +(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... -n*5)$

Which is divisible by $5$.

Do the same for $4^n + 5^n + 6^n = (3+1)^n + (6-1)^n + 6^n$ to show it is divisible by $3$.

fleablood
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  • When you wrote $41526^{2k-1},$ did you mean $41526^{2k-\color{red}2}$? – J. W. Tanner Nov 21 '19 at 11:49
  • Almost certainly, I did. The $20$ provides a $5$ so $20 = 45$ and the $6^{2k-1}$ provides a $3$ so $6^{2k-1} = 326^{2k-2}$ and $206^{2k-1} = 4152*6^{2k-2}$. – fleablood Nov 21 '19 at 14:20
  • When you say $[16(4^{2−1}+5^{2−1}+6^{2−1})]$ = [16∗15], what do you mean there? What is $N$? – lswift Nov 13 '22 at 11:54
  • $N$ is whatever it needs to be to meake $4^{2k-1} + 5^{2k-1}+6^{2k-1} = 15N$. In other words $N =\frac {4^{2k-1} + 5^{2k-1} +6^{2k-1}}{15}$. See the original post. We are assuming in an induction step that $4^{2k-1} + 5^{2k-1} + 5^{2k-1}$ is divisible by $15$. That means $4^{2k-1} + 5^{2k-1} + 5^{2k-1} = 15 SOMETHING$. $N$ is that $SOMETHING$. – fleablood Nov 13 '22 at 16:20
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Hint : \begin{eqnarray*} 31(4^n+5^n+6^n)= 4^{n+2}+5^{n+2}+6^{n+2} +15 \times 4^n+ 6 \times 5^n - 5 \times 6^n. \end{eqnarray*}

Donald Splutterwit
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