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 February 10, 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, a_n+1=2a_n+3b_n, b_n+1=a_n+2b_n.)

For n $\ge$ 1 we have b_n>1.

Statistics:

7 students sent a solution.

5 points: Ágoston Tamás, Gyarmati Máté, Janzer Olivér, Mester Márton, Omer Cerrahoglu, Strenner Péter, Szabó 789 Barnabás.

Problems in Mathematics of KöMaL, January 2012

7 students sent a solution.
5 points:	Ágoston Tamás, Gyarmati Máté, Janzer Olivér, Mester Márton, Omer Cerrahoglu, Strenner Péter, Szabó 789 Barnabás.

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