How con I Show that $29^{41}+41^{29}$ is divisible by 35. attempt: I think it shouldn’t be difficult but not sure how to prove it.
29 is not a multiple of 5 nor of 7 same for 41.
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Hint: what are $29,41\pmod {35}$? – lulu Jun 14 '21 at 21:06
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1What have you tried - you've tagged it modular arithmetic ... – Mark Bennet Jun 14 '21 at 21:06
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$29^{41}+41^{29} \equiv -1+1\equiv 0 \pmod{5}$ ; $29^{41}+41^{29} \equiv 1 - 1 \equiv 0 \pmod{7}$ therefore $29^{41}+41^{29} \equiv 0 \pmod{35}$ – Evariste Jun 14 '21 at 21:07
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Hint: Use the Chinese Remainder theorem and Euler-Fermat's theorem. – Bernard Jun 14 '21 at 21:10
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Hint: $,J,K$ odd $\Rightarrow (-6)^{J}+ 6^{K}\equiv 0\pmod{5\ &\ 7}\ $ so also $\bmod 5\cdot 7\ \ $ – Bill Dubuque Jun 14 '21 at 21:12
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Hint: $,J,K,$ odd $\Rightarrow (-a)^{J}+ a^{K}\equiv 0\pmod{a!-!1\ &\ a!+!1}\ $ so also $\bmod a^2-1$ when $,2\mid a,,$ since then ${\rm lcm}(a!-!1,a!+!1) = (a!-!1)(a!+!1),$ by $\gcd(a!-!1,a!+!1) = \gcd(a!-!1,2) = 1\ \ $ – Bill Dubuque Jun 14 '21 at 21:20
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Trivial congruence arithmetic using congruence laws shows $,x\equiv 0,$ both mod $5$ & $7$ therefore $,x\equiv 0,$ mod their lcm $= 5\cdot 7,$ by CCRT in the linked dupe. – Bill Dubuque Jun 14 '21 at 21:39
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Alternatively use $,6^2\equiv 1\pmod{35},$ and mod order reduction (cf. 2nd dupe). Such nontrivial sqrts of $1$ always correspond to factorizations of the modulus into coprime factors. – Bill Dubuque Jun 14 '21 at 21:52
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The diract way:
$29^{41} + 41^{29}$ mod $35$ = $(-6)^{41} + 6^{29}$ mod $35$ = $-6 \cdot (-6)^{40} + 6\cdot6^{28}$ mod $35$ = $-6 \cdot 36 ^{20} + 6 \cdot 36^{14}$ mod $35$ = $- 6 + 6$ mod $35$ = $0$ mod $35$
So we can see that $35$ divides $29^{41} + 41^{29}$.
Uroš Kosmač
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direct* additive inverses raised to odd powers being multiplicative inverses doesn't quite help here sadly ... – Roddy MacPhee Jun 19 '21 at 01:01