Algebra 361wherek=n− 21. Sincen≥5, both factors at the numerator are greater than 5, which
shows that after canceling the denominator, the expression on the right can still be written
as a product of two numbers. This proves thatNis not prime.
(T. Andreescu, published in T. Andreescu, D. Andrica, 360Problems for Mathematical
Contests, GIL, 2003)
86.We use the identity
a^5 − 1 =(a− 1 )(a^4 +a^3 +a^2 +a+ 1 )applied fora= 5397. The difficult part is to factora^4 +a^3 +a^2 +a+1. Note that
a^4 +a^3 +a^2 +a+ 1 =(a^2 + 3 a+ 1 )^2 − 5 a(a+ 1 )^2.Hence
a^4 +a^3 +a^2 +a+ 1 =(a^2 + 3 a+ 1 )^2 − 5398 (a+ 1 )^2
=(a^2 + 3 a+ 1 )^2 −( 5199 (a+ 1 ))^2
=(a^2 + 3 a+ 1 + 5199 (a+ 1 ))(a^2 + 3 a+ 1 − 5199 (a+ 1 )).It is obvious thata−1 anda^2 + 3 a+ 1 + 5199 (a+ 1 )are both greater than 5^100. As for
the third factor, we have
a^2 + 3 a+ 1 − 5199 (a+ 1 )=a(a− 5199 )+ 3 a− 5199 + 1 ≥a+ 0 + 1 ≥ 5100.Hence the conclusion.
(proposed by Russia for the 26th International Mathematical Olympiad, 1985)
87.The number from the statement is equal toa^4 +a^3 +a^2 +a+1, wherea= 525 .As
in the case of the previous problem, we rely on the identity
a^4 +a^3 +a^2 +a+ 1 =(a^2 + 3 a+ 1 )^2 − 5 a(a+ 1 )^2 ,and factor our number as follows:
a^4 +a^3 +a^2 +a+ 1 =(a^2 + 3 a+ 1 )^2 −( 513 (a+ 1 ))^2
=(a^2 + 3 a+ 1 + 513 (a+ 1 ))(a^2 +a+ 1 − 513 (a+ 1 )).The first factor is obviously greater than 1. The second factor is also greater than 1, since
a^2 +a+ 1 − 513 a− 513 =a(a− 513 )+(a− 513 )+ 1 ,anda> 513. This proves that the number from the statement of the problem is not prime.
(proposed by Korea for the 33rd International Mathematical Olympiad, 1992)