Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
 Already signed up? New to KöMaL?

# 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