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2013 JBMO Problems/Problem 1

Revision as of 14:27, 20 October 2022 by Satisfiedmagma (talk | contribs) (→‎Solution)

Problem

Find all ordered pairs $(a,b)$ of positive integers for which the numbers $\dfrac{a^3b-1}{a+1}$ and $\dfrac{b^3a+1}{b-1}$ are both positive integers


Solution

Adding $1$ to both the given numbers we get:

$\dfrac{a^3b-1}{a+1} + 1$ is also a positive integer so we have:

$\dfrac{a^3b+a}{a+1}$ = $\dfrac{a(a^2b+1)}{a+1}$ is a positive integer

$\implies (a+1) \mid (a^2b+1)$ $\implies (a+1) \mid (((a+1) - 1)^2b+1)$ $\implies (a+1) \mid (b+1)$

Similarly,

$\dfrac{b^3a+1}{b-1} + 1$ is also a positive integer so we have:

$\dfrac{b^3a+b}{b-1}$ = $\dfrac{b(b^2a+1)}{b-1}$ is a positive integer

$\implies (b-1) | (b^2a+1)$ $\implies (b-1) | (((b-1) + 1)^2a+1)$ $\implies (b-1) | (a+1)$

Combining above $2$ results we get:

$(b-1) | (b+1)$

$\implies b=2,3$

$Case 1: b=2$ $\implies a+1|3 \implies a=2$ which is a valid solution.

$Case 2: b=3$ $\implies a+1|4 \implies a=1,3$ which are valid solutions.

Thus, all solutions are: $(2,2), (1,3), (3,3)$


$Kris17$

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