Determine all the triples of positive integers $a,b,c$ such that $a^7+b^7=7^c$.
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4What do terns have to do with it? – barak manos Jan 28 '15 at 13:05
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1Welcome to MathSE. When you pose a question here, you should explain its source, include any work you have done, and explain why you are stuck so that we can write answers appropriate to your skill level. – N. F. Taussig Jan 28 '15 at 13:23
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my bad, i was thinking in italian sorry; – charlizard Jan 28 '15 at 13:24
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No problem, using other languages is encouraged, but this way it may be a little confusing. – Bart Michels Jan 28 '15 at 13:26
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I was already well advised by N.F.Taussig and i apologize; you think it would make any sense if i re-edit the question after i got answered? maybe yes. Any way next question i will provide more as Taussig was suggesting me. – charlizard Feb 11 '15 at 14:13
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See here for the more general $x^p + y^p = p^z$ – Bart Michels Apr 02 '15 at 16:16
1 Answers
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Zsigmondy's theorem is the way to go. Applying Zsigmondy shows that $a^7+b^7$ has at least $2$ distinct prime divisors (one appearing in the factorisation of $a+b$, the other given by Zsigmondy), hence it can not be a power of $7$.
Of course, Zsigmondy is a bit an overkill here.
A more elementary approach would be to use the result of this question (or imitating the proof of the result given there, which isn't too hard): $$\gcd\left(\frac{a^7+b^7}{a+b},a+b\right)=\gcd(7\gcd(a,b)^6,a+b).$$
If (without loss of generality) $\gcd(a,b)=1$, this would mean that either
- $a+b=7$, which leaves only a few cases to check
- $a+b>7$, in which case the above shows that $\frac{a^7+b^7}{a+b}=7$, hence $a^7+b^7=7(a+b)$ which is impossible if at least one of $a,b$ is $\geqslant2$.
Bart Michels
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Either way, I think an approach would be similar to the one in the linked question (binomial expansion and investigating divisibility by powers of $7$ of each of the individual terms) or use some deeper properties of cyclotomic polynomials (as Zsigmondy does). I thought it would be interesting to present some more general results here, because these can be used to solve any equation of this type. – Bart Michels Jan 28 '15 at 13:49
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Note that the second technique does not require FTA nor modular arithmetic. – Bart Michels Jan 28 '15 at 13:56