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Difference between revisions of "2013 JBMO Problems/Problem 1"

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<math>Kris17</math>
 
<math>Kris17</math>
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== Solution 1.5 (credit to dskull16) ==
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To get the two results:
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<math> (a+1) \mid (b+1)</math>
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<math> (b-1) \mid (a+1)</math>
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We can also add zero to the numerator as follows:
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<math>\dfrac{a^3b-1}{a+1} = \dfrac{a^3b + b - (b+1)}{a+1}</math>
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<math>\implies (a+1) \mid (b+1)</math>
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since <math>a^3 + b = b(a+1)(a^2-a+1)</math>
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<math>\dfrac{b^3a+1}{b-1} = \dfrac{b^3a - a + (a+1)}{b-1}</math>
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<math>\implies (b-1) \mid (a+1)</math>
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since <math>b^3a - a = a(b-1)(b^2 + b + 1)</math>
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Then proceed as above.

Latest revision as of 17:09, 10 March 2024

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: $(2,2), (1,3), (3,3)$


$Kris17$

Solution 1.5 (credit to dskull16)

To get the two results: $(a+1) \mid (b+1)$ $(b-1) \mid (a+1)$

We can also add zero to the numerator as follows:

$\dfrac{a^3b-1}{a+1} = \dfrac{a^3b + b - (b+1)}{a+1}$ $\implies (a+1) \mid (b+1)$

since $a^3 + b = b(a+1)(a^2-a+1)$


$\dfrac{b^3a+1}{b-1} = \dfrac{b^3a - a + (a+1)}{b-1}$ $\implies (b-1) \mid (a+1)$

since $b^3a - a = a(b-1)(b^2 + b + 1)$

Then proceed as above.

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