I tried by induction but at induction step I struck.I am bad at proving statements which involve variables as exponents.I gladly appreciate any help
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6$a+1 \mid a^3+1$ – user26857 Oct 20 '19 at 10:27
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1Possible duplicate of factorization of a^n+1? As usual, some may feel that a less general duplicate target is required. – Jyrki Lahtonen Oct 20 '19 at 12:20
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Would you like to accept an answer, Mathematical? – Gerry Myerson Dec 04 '19 at 21:22
4 Answers
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$8^n+1=(2^n+1)\left((2^n)^2-2^n+1\right)$.
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Use modular arithmetic: $2^n\equiv -1 \mod 2^n+1$,
so $8^n=2^{3n}=(2^n)^3\equiv(-1)^3=-1\mod 2^n+1$
J. W. Tanner
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For $n=1, 2^n+1= 2+1= 3$ which divides $(8+1)=9$.
So let it be true for $n = k$ where $k \in \mathbb{N}$ . This means $2^k+1 | 8^k+1$.
Now for $n = k+1$ we get, $$ 8^{k+1} +1 = (4^{k+1})(2^{k+1}+1)-(2^{k+1})(2^{k+1}+1)+(2^{k+1}+1) = (2^{k+1}+1)(4^{k+1}-2^{k+1}+1)$$
which is a multiple of $2^{k+1}+1$.
Hence by the principle of mathematical induction it is proved that $(2^n)+1 $ divides $(8^n)+1$ where $n \in \mathbb{N}$. This is a proof using induction.
Feng
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Shiva 1729
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This induction proof is not good as when you prove $n=k+1$ is correct, you do not use the information of when $n=k$. So, this is not a great induction proof. – MafPrivate Oct 20 '19 at 11:09
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But that information is not necessary to use in this case and I think it is better to do this problem without using induction. – Shiva 1729 Oct 21 '19 at 11:41
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As $8^n=\left(2^3\right)^n=2^{3n}=\left(2^n\right)^3$, let $a=2^n$: $$n^3+1=(n+1)(n^2-n+1)\\\therefore n+1|n^3+1$$
MafPrivate
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