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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0@Num: I can't think of any other interpretation without the brackets. I don't see any ambiguity there. Also, my guess is Amir is just posting challenge problems and this is not really homework. – 2011-04-08
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0@Moron As far as parentheses, fine. As far as homework, I agree. But, it'd still be nice to know what he tried. – 2011-04-08
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3@NUm: Yes, not only that. The source seems to missing. And I am guessing he already knows the solution. (I was about to write a comment asking for that info...) – 2011-04-08
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0If this is so, then $a$ will be strictly less than $b$, and $a^2+b$ will divide $ab-1$. – 2011-04-08
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0I have a proof, but I gotta sleep now. – 2011-04-08
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0This implies that there is some integer $k$ such that $(2k)^2$ divides $b^2 -(2bk)^2 -4k$; hence a and b cannot be primes. I hope all cases can be reduced to primes, but ... – 2011-04-08
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0@awllower Actually, a=b works just fine and is always a solution. So, now you can assume that a is strictly less than b as you move on. – 2011-04-08
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1@Moron @Numth This posts are starting to look a lot like Chandru1's posts in the past. – 2011-04-08
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1@Adrian: Very true... I guess we can point Amir to the meta threads present already. – 2011-04-08
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1@Moron Yes, I think that would be a good idea. – 2011-04-08
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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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0I think that sequence is just $\tau(a^4+a)/2$, half the number of factors of $a^4+a.$ – 2011-04-08
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0Whoops, wrong, it's equal to the number of factors of $a^4+a$ that are less than $a^2-a+1$. This might often equal my result, but not always. – 2011-04-08
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1Okay, I think that there is a way to prove that there are no factors of $a^4+a$ greater than $a^2-a+1$ and less than $\sqrt{a^4+a}, so my original count, $\frac{\tau(a^4+a)}{2}$, should be correct – 2011-04-08
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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