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# We will show by induction that for any positive integer n, 37n−1+ 57n−1−1 ≡ 0 (mod 7n)

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Problem 11992

(American Mathematical Monthly, Vol.124, August-September 2017) Proposed by N. Safaei (Iran).

Prove that, for every positive integer n, there is a positive integer m such that3m+5m−1 is divisible by7n.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. We will show by induction that for any positive integer n, 37n−1+ 57n−1−1 ≡ 0 (mod 7n).

It trivially holds for n = 1. Now we assume that the congruence holds for some n ≥ 1. Then there is an integer q such that 57n−1= 1 − x + q7n where x = 37n−1, and

37n+ 57n−1 = x7+ (1 − x + q7n)7−1

= x7+ (1 − x)7−1 + 7(q7n)(1 − x)6+

7

X

k=2

7 k



(q7n)k(1 − x)7−k

≡x7+ (1 − x)7−1 ≡ 7x(x − 1)(x2−x+ 1)2≡0 (mod 7n+1) where the last step holds as soon as we show that x2−x+ 1 ≡ 0 (mod 7n).

Note that, since ϕ(7n) = 6 · 7n−1, then, by the Fermat’s little theorem,

(x − 1)(x + 1)(x2+ x + 1)(x2−x+ 1) = x6−1 = 3ϕ(7n)−1 ≡ 0 (mod 7n).

Moreover x ≡ 3 (mod 7), hence 7 does not divide (x − 1)(x + 1)(x2+ x + 1) ≡ 6 (mod 7), which

implies that 7ndivides x2−x+ 1 as claimed. 

Riferimenti

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Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma,

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma,

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma,

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma,

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Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma,

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