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Problem A. 551. (January 2012)

A. 551. Show that there exist infinitely many pairs (a,b) of positive integers with the property that a+b divides ab+1, a-b divides ab-1, b>1 and a>b\sqrt3-1.

(5 pont)

Deadline expired on 10 February 2012.


Solution (outline). In view of the solution of Problem A. 545., we look for positive integers a,b with the property

a2-3b2=-2.(1)

If some pair (a,b) of positive integers satisfies (1) then a and b have the same parity,

ab+1 = ab+\frac{3b^2-a^2}2 = (a+b)\frac{3b-a}2, so a+b|ab+1,

ab-1 = ab-\frac{3b^2-a^2}2 = (a-b)\frac{a+3b}2, so a-b|ab-1;

finally

\sqrt3 b = \sqrt{a^2+2} < \sqrt{a^2+2a+1} = a+1, so a>\sqrt3b-1.

The Pell-like equation (1) has infinitely many solutions which can be represented as


a_n\pm b_n\sqrt3 = \big(1\pm \sqrt3\big)\cdot\big(2\pm \sqrt3\big)^n,


\matrix{
a_n = \dfrac{
\big(1+\sqrt3\big)\cdot\big(2+\sqrt3\big)^n+
\big(1-\sqrt3\big)\cdot\big(2-\sqrt3\big)^n}2, & \cr
b_n = \dfrac{
\big(1+\sqrt3\big)\cdot\big(2+\sqrt3\big)^n-
\big(1-\sqrt3\big)\cdot\big(2-\sqrt3\big)^n}{2\sqrt3}
& (n=0,1,2,\ldots).}

(The same pairs can be generated by the recurrence a0=b0=1, an+1=2an+3bn, bn+1=an+2bn.)

For n\ge1 we have bn>1.


Statistics:

8 students sent a solution.
5 points:Ágoston Tamás, Gyarmati Máté, Janzer Olivér, Mester Márton, Omer Cerrahoglu, Strenner Péter, Szabó 789 Barnabás.
Unfair, not evaluated:1 solution.

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