I factored this out to be $(7^n-1)(7^n+1)(7^{2n}+1)$, and I know I need to prove that I can always factor out a 3 and a 5, but I have no idea how to get there, especially without induction. My professor said I can use the rule of three consecutive integers (how one of them will always be a multiple of 3) to get started, but that doesn't look to be possible here.
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I suppose $7^{4n}-1=2401^n-1=(2401-1)(2401^{n-1}+2401^{n-2}+\cdots+2401+1)$, but really can you prove $x^n-1=(x-1)(x^{n-1}+\cdots+1)$ "without induction"? – Angina Seng Jun 07 '20 at 05:33
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I could prove it by induction, but I'm trying to get full credit on this assignment – Michael Johnston Jun 07 '20 at 05:36
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The site has a strict policy against homework problems – Arjun Jun 07 '20 at 07:58
3 Answers
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$$(7^{4})^n-1=(2401)^n-1=(2401-1)P(n)=2400 P_{n-1}(2401)$$ which is divisible by 15. Here $P_{n-1}(2401)$ is a polynomial of degree $n-1$: $1+(2301)+(2401)^2+....+(2401)^{n-1}$ - an integer.
Z Ahmed
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@Arjun Rana Actually this is a polynomial of (2401) of degree $n-1$ like : 1+(2301)+(2401)^2+....+(2401)^{n-1}$ which is a whole number..Please see the edit. – Z Ahmed Jun 07 '20 at 09:28
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I am not sure if you are allowed to use or know modular arithmetic, but we know that $7^4=2401$ which has a remainder of $1$ when divided by $15$. Then $7^{4n}-1= (7^4)^n-1=(2401)^n-1$ has remainder $1^n-1=1-1=0$ when divided by $15$; that is, $7^{4n}-1$ is divisible by $15$. In modular arithmetic language, this is because $$ 7^{4n}-1 \equiv (2401)^n-1 \equiv 1^n-1 \equiv 0 \mod 15 $$
mathematics2x2life
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A simple and direct way is using the binomial formula in the form
- ($\star$): $(ak+b)^m = const\cdot k + b^m$
So, you get
- Divisibility by $3$: $$7^{4n}-1 = \underbrace{(2\cdot 3 + 1)^{4n}}_{\stackrel{(\star)}{=}3K+1} - 1 = 3K$$
- Divisibility by $5$: $$7^{4n}-1 = \underbrace{(50 - 1)^{2n}}_{\stackrel{(\star)}{=}5L+1} - 1 = 5L$$
trancelocation
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