How can one find all positive integers $a,b$ such that $a^2+b \mid b^2+a$?
All positive integers $a,b$ such that $a^2+b \mid b^2+a$
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number-theory
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1Here are$a$couple of threads about asking questions whose answers you already know/asking puzzles: [Questions whose answers are known to OP](http://meta.math.stackexchange.com/questions/1839/questions-whose-answers-are-known-to-the-op) and [Are puzzles on-topic](http://meta.math.stackexchange.com/questions/1190/is-math-puzzle-on-topic-or-off-topic). Please also see [How to ask a Good question](http://meta.math.stackexchange.com/questions/588/how-can-i-ask-a-good-question) – 2011-04-08
1 Answers
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First of all, as already noted, $a\le b$. Next, since $b^2+a=(b-a^2)(b+a^2)+a+a^4$, it follows that $a^2+b$ must divide $a^4+a$. We see then that given $a\ge1$, if a positive integer $b$ is such that $a^2+b\mid b^2+a$, then $b=d-a^2$, where $d$ is a divisor of $a^4+a$ greater than $a^2$. For instance, $b=a^4-a^2+a$ is a solution for all $a\in\mathbb{N}$.
A one-liner in Mathematica to compute all the $b$'s for a given $a$ is
f[a_] := Select[Divisors[a + a^4]-a^2, Positive]
Some examples:
a=2 ; b -> {2,5,14} a=3 ; b -> {3,5,12,19,33,75} a=4 ; b -> {4,10,36,49,114,244} a=5 ; b -> {5,10,17,20,38,45,65,80,101,185,290,605} a=6 ; b -> {6,26,57,150,181,398,615,1266} a=7 ; b -> {7,37,123,252,295,553,1155,2359} a=8 ; b -> {8,12,44,50,88,107,152,164,278,392,449,620,962,1304,1988,4040} a=9 ; b -> {9,65,138,284,357,576,649,1014,1233,2109,3204,6489} a=10 ; b -> {10,30,43,54,82,186,285,355,615,670,810,901,1330,1902,4905,9910}
One obvious question to ask is ¿how many solutions are there for a given $a$? Here are the first numbers in the sequence:
Table[Length[f[a]], {a, 2, 100}] {3, 6, 6, 12, 8, 8, 16, 12, 16, 18, 24, 8, 24, 20, 10, 32, 12, 24, 36, 16, 8, 36, 48, 12, 40, 48, 12, 24, 32, 36, 36, 32, 16, 48, 36, 16, 48, 32, 32, 24, 16, 24, 48, 48, 16, 60, 60, 36, 72, 24, 24, 20, 64, 32, 96, 32, 8, 72, 24, 16, 64, 42, 56, 96, 32, 12, 108, 96, 16, 32, 24, 16, 36, 144, 24, 48, 16, 20, 180, 20, 32, 36, 96, 32, 24, 64, 64, 64, 48, 24, 36, 32, 32, 144, 48, 24, 48, 108, 18}
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0It's simpler to work with the conjugates $\rm\ b' = k - b = a\:(1-ab)/(a+b^2)\:.\:$ E.g. for $\rm\ a = 7\ $ we have $\rm\: -b' = \{6, 21, 35, 41, 42, 45, 47, 48 = a^2-1\}$ versus the more complicated $\rm\ \ \ b\ =\: \{7,37,123,252,295,553,1155,2359\}\:.$ – 2011-04-08