Mathematical and Physical Journal
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Problem B. 4421. (January 2012)

B. 4421. Let t denote a fixed integer. Show that for every odd prime number p, there exists a positive integer n, such that (3-7t)2n+(18t-9)3n+(6-10t)4n is divisible by p.

(Suggested by K. Kalina, Budapest)

(5 pont)

Deadline expired on February 10, 2012.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A \(\displaystyle p=3\) esetben nyilván \(\displaystyle n=1\) megfelelő lesz. Ha \(\displaystyle p\ge 5\) akkor a kis Fermat-tétel szerint \(\displaystyle 2^{p-1}, 3^{p-1}\) és \(\displaystyle 4^{p-1}\) is 1 maradékot ad \(\displaystyle p\)-vel osztva, vagyis

\(\displaystyle 2^{p-1}=Ap+1, \quad 3^{p-1}=Bp+1, \quad 4^{p-1}=Cp+1\)

írható alkalmas \(\displaystyle A,B,C\) egész számokkal. Ekkor \(\displaystyle n=p-2\) esetén

\(\displaystyle (3-7t)2^n+(18t-9)3^n+(6-10t)4^n= \frac{3-7t}{2}2^{p-1}+\frac{18t-9}{3}3^{p-1}+\frac{6-10t}{4}4^{p-1}=\)

\(\displaystyle =\frac{3-7t}{2}Ap+\frac{18t-9}{3}Bp+\frac{6-10t}{4}Cp= \frac{\left(6A(3-7t)+4B(18t-9)+3C(6-10t)\right)p}{12}.\)

Ez az egész szám nyilván osztható \(\displaystyle p\)-vel.


15 students sent a solution.
5 points:Ágoston Tamás, Barna István, Di Giovanni Márk, Fehér Zsombor, Gyarmati Máté, Janzer Olivér, Kiss 902 Melinda Flóra, Maga Balázs, Mester Márton, Nagy Róbert, Schultz Vera Magdolna, Strenner Péter, Szabó 789 Barnabás.
3 points:1 student.
2 points:1 student.

Problems in Mathematics of KöMaL, January 2012