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Problem A. 622. (September 2014)

A. 622. Prove that \(\displaystyle \dfrac{7^{7^{k+1}}+1}{7^{7^{k}}+1}\) is a composite number for every nonnegative integer \(\displaystyle k\).

(5 pont)

Deadline expired on October 10, 2014.


Solution. Notice that

\(\displaystyle \dfrac{x^7+1}{x+1} = x^6-x^5+x^4-x^3+x^2-x+1 = (x+1)^6 -7x \cdot (x^2+x+1)^2. \)

If \(\displaystyle x=7^{7^k}\) then \(\displaystyle 7x=7^{7^k+1}\) is a square, so \(\displaystyle \dfrac{7^{7^{k+1}}+1}{7^{7^{k}}+1}\) is the difference between two squares. Hence,

\(\displaystyle \dfrac{7^{7^{k+1}}+1}{7^{7^{k}}+1} = \bigg( \Big(7^{7^k}+1\Big)^3 + 7^{\frac{7^{k}+1}2}\Big(7^{2\cdot 7^k}+7^{7^k}+1\Big) \bigg) \bigg( \Big(7^{7^k}+1\Big)^3 - 7^{\frac{7^{k}+1}2}\Big(7^{2\cdot 7^k}+7^{7^k}+1\Big) \bigg). \)

The second factor is greater than \(\displaystyle 1\), because

\(\displaystyle \Big(7^{7^k}+1\Big)^3 - 7^{\frac{7^{k}+1}2}\Big(7^{2\cdot 7^k}+7^{7^k}+1\Big) \ge \Big(7^{7^k}+1\Big)^3 - 7^{7^{k}}\Big(7^{2\cdot 7^k}+7^{7^k}+1\Big) = 2\cdot7^{2\cdot 7^k} +2\cdot7^{7^k} +1 > 1. \)


Statistics:

8 students sent a solution.
5 points:Adnan Ali, Fehér Zsombor, Saranesh Prembabu, Shapi Topor, Szabó 789 Barnabás.
4 points:Ahaan S. Rungta.
0 point:2 students.

Problems in Mathematics of KöMaL, September 2014