How to prove that for any natural number $n$, $10|(3^{2+4n}+1)$ using induction?

Question

I've seen a few answers proving it without induction, using congruence modulo but we haven't actually covered that yet so I don't think that's how I'm supposed to answer the question.
I understand why it's true, $3^{2+4n}=3^2*3^{4n}$ and $3^{4n}$ will always give a number ending in 1, it gets multiplied by $3^2=9$ to give a number ending in 9 and adding the $1$ will always give a multiple of 10, I just don't know how to prove it, especially without using congruence modulo.

So far I have

Base case: $n=1$ : $3^{2+4}+1=730$ , $10|730$ so $P(1)$ is true.

Inductive step: We assume that $n=k$ is true, that is $10|(3^{2+4k}+1)$. We must now show that n=k+1 is true, that is $10|(3^{6+4k}+1)$

$3^{6+4k}+1=3^4*3^{2+4k}+1$

But I'm not sure where to go from there. Any help is appreciated!

Answer 1

Induction isn't too bad:

Let $a_n=3^{2+4n}+1$. Inductively, suppose that $10\,|\,a_{n-1}$.

Then: $$a_n=81\times (a_{n-1}-1)+1=81\times a_{n-1}-81+1=81\times a_{n-1}-80$$ and we are done.

Answer 2

Write $9^{1+2n}+1=3^{2+4n}+1 = 10k$. If we multiply this with $9^2$ we get

$$ 810k = 9^2\cdot (9^{1+2n}+1) = 9^{1+2(n+1)}+81 = 9^{1+2(n+1)}+1+80$$

So $$9^{1+2(n+1)}+1= 810k-80 = 10(...)$$

and we are done.

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You can avoid induction:

$$3^{2+4n}+1 =(3^2)^{1+2n}+1 $$ $$=9^{1+2n}+1 \equiv _{10} (-1)^{2n+1}+1 \equiv _{10} -1+1 \equiv _{10}0$$

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Answer 3

For $n=1$ we get $$10|(3^{6}+1)= 730$$

Assume that $$10|(3^{2+4n}+1)$$

Then $$3^{2+4(n+1)}+1 = 3^{2+4n} (81) +1 = (3^{2+4n}+1-1) (81) +1 $$

$$=(3^{2+4n}+1)(81) -81 +1 = (3^{2+4n}+1)(81) -80 $$

Which is a multiple of $10$

Answer 4

$P(1)$ is true

Assume $p(n)$ is true

$P(n+1)= 3^{4} p(n)-80$

$P(n)$ is true thus $p(n+1)$.

Answer 5

Hint $ $ An easy induction shows $81^n$ has units digit $1$ so $\,\overbrace{9(81)^n\!+1}^{\large 3^{\LARGE 2+4n}+\ 1}$ has units digit $9(1)+1\! =\! 0$.

For the inductive step use: $ $ if $a,b$ have units digit $1$ then so does their product.

Remark $ $ If you know modular arithmetic then it boils down to the following

$\bmod 10\!:\,\ a\equiv 1\,\Rightarrow\, a^n\equiv 1,\,$ with inductive step $\,a^{n+1}\equiv a(a^n)\equiv 1(1)\equiv 1$

That's simply the inductive proof of the Congruence Power Rule using the Product Rule. So the induction in the OP is just a very special case of the inductive proof of the Power Rule, and using modular arithmetic clarifies the arithmetical essence of the matter.