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Problem A. 545. (November 2011)

A. 545. Prove that whenever a>b>1 are integers such that a+b divides ab+1 and a-b divides ab-1 then a<b\sqrt3.

(Kolmogorov Cup, 2009)

(5 pont)

Deadline expired on December 12, 2011.

Solution. Since b2-1=b(a+b)-(ab+1)\equiv0 (mod  (a+b)) and b2-1=-b(a-b)+(ab-1)\equiv0 (mod  (a-b)), both a+b and a-b divide (b2-1). Hence, b2-1 is divisible by the least common multiple of a+b and a-b. By the conditions b2-1 and a-b are positive, so this implies


Next we show that a and b are coprime; moreover, the greatest common divisor of a+b and a-b is at most 2. Let d=(a,b). Then d|a|ab and d|a+b|ab+1, so d|(ab+1)-ab=1 and thus d=1. Now let e=(a+b,a-b). Then by e|(a+b)+(a-b)=2a and e|(a+b)-(a-b)=2b we have e|(2a,2b)=2(a,b)=2. Therefore, e\le2.

From the identity [x,y]=\frac{xy}{(x.y)} we get

 [a+b,a-b] = \frac{(a+b)(a-b)}{(a+b,a-b)} \ge \frac{a^2-b^2}{2}.

Combining this with (1), we get

\frac{a^2-b^2}2 \le b^2-1


a < b\sqrt3.


9 students sent a solution.
5 points:Ágoston Tamás, Gyarmati Máté, Janzer Olivér, Kovács 444 Áron, Mester Márton, Omer Cerrahoglu, Strenner Péter, Szabó 928 Attila.
0 point:1 student.

Problems in Mathematics of KöMaL, November 2011