Problem C. 940. (April 2008)

C. 940. Prove if n is a positive integer then 2⁴ⁿ-1 or 2⁴ⁿ+1 is divisible by 17.

(5 pont)

Deadline expired on May 15, 2008.

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

Megoldás: 2⁴ⁿ=(2⁴)ⁿ=16ⁿ.

Legyen n páros, ekkor n=2k, és 2⁴ⁿ-1=16^2k-1^2k=(16+1)(16^2k-1-16^2k-2+16^2k-3-...+16-1)=17^.a, ahol a $\geq$ 1 és egész, amennyiben k $\geq$ 1. Vagyis ha n pozitív páros szám, akkor 17|2⁴ⁿ-1.

Legyen n páratlan, ekkor n=2k+1, és 2⁴ⁿ+1=16^2k+1-1^2k+1=(16+1)(16^2k-16^2k-1+16^2k-2-...-16+1)=17^.b, ahol b $\geq$ 1 és egész, amennyiben k $\geq$ 1. Vagyis ha n pozitív páratlan szám, akkor 17|2⁴ⁿ+1.

Statistics:

172 students sent a solution.

5 points: 67 students.

4 points: 76 students.

3 points: 16 students.

2 points: 2 students.

1 point: 3 students.

0 point: 8 students.

Problems in Mathematics of KöMaL, April 2008

172 students sent a solution.
5 points:	67 students.
4 points:	76 students.
3 points:	16 students.
2 points:	2 students.
1 point:	3 students.
0 point:	8 students.

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