Find all $m \in \mathbb{Z}$, for which the expression $7^m - 3^m$ a prime. What would be the general strategy for solving this? I have no clue where to begin. I tried picking $m$ and checking some values but I don't see any relation between them. This question can be solved using modular arithmetic as well but im not sure how. What I do know is that $7^m - 3^m > 1$.
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1For a start, when $m$ is odd, $7^m-3^m$ is even. – saulspatz Mar 11 '21 at 02:18
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1Reduce mod $4$. – Mark Saving Mar 11 '21 at 02:19
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1$7^m - 3^m$ is always even... – Tob Ernack Mar 11 '21 at 02:20
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Use the Factor Theorem as in the linked dupe. – Bill Dubuque Mar 11 '21 at 08:49
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But $7^m - 3^m$ for positive $m$ is not only even but is a multiple of $4$ which is not prime [ETA as noted in comments already]. So definitely not prime.
In general, if $a$, $b$, and $c$ are positive integers satisfying $a-b = c$, then working mod $c$, one can see that $a^m-b^m$ is always a multiple of $c$.
Mike
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We have $7^{m}-3^{m}$ and m is positive. So, $$7^{m}-3^{m}=(7-3)(7^{m-1}+7^{m-2}.3+7^{m-3}.3^{2}+......7^{2}.3^{m-3}+7.3^{m-2}+3^{m-1})$$ $$7^{m}-3^{m}=\mathbf{4}.(7^{m-1}+7^{m-2}.3+7^{m-3}.3^{2}+......7^{2}.3^{m-3}+7.3^{m-2}+3^{m-1})$$ $$7^{m}-3^{m}=\mathbf{4}.g(m)$$ So 4 is a root of this for m ($m\in \mathbb{Z^{+}}$) positive. So $7^{m}-3^{m}$ is not a prime.
Tharindu Gimras
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1where can i read more about the factorization you came up with? Im not sure how you split into that factorization. Is it just the regular factorization of $a^n - b^n$? Im just confused about the "." symbols used in the factorization – UnKnoWnZ Mar 11 '21 at 02:31
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@UnKnoWnZ I just saw this in x^2-y^2=(x-y)(x+y) , x^3-y^3=(x-y)(x^2+x.y+y^2) , x^4-y^4=(x-y)(x^3+x^2.y+x.y^2+y^3) format. Then I solved this like this.I learned this from my mathematics teacher in the past times. So if you need more information about this factorisation, just go to link. – Tharindu Gimras Mar 11 '21 at 02:43