I wanted to find pair of (integers) numbers (actually just 3 of them) such that:

$$9U=9K_1 + 4 K_1^2$$

I've tried choosing a fixed integer starting at 1 for U and then had mathematica/wolfram find $K_1$. So far I've tried up to 6 with no luck. However, I wasn't sure if this even had a solution and my knowledge in number theory is quite limited. So I thought it would be wise to ask the expert community on this.

Just in case this is impossible to solve, I'd be sort of ok with numbers such that $U,K_1$ are at least $ \epsilon < 0.5$ close to an integer. This is not the ideal case but if its impossible I accept approximations or I need to change my conditions to make this problem solvable.

If you are curious this equation came from trying to find integers that satisfy:

$$ 9U = 3K_1 + 2 K_1 K_2 + 3 K_2 $$

for some nice properties of my problem choosing $4K_1 = 2K_2 \iff 2K_1 = K_2$. Though I guess we could change that condition if necessary but it wouldn't be ideal.

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    $\begingroup$ clearly there is exactly one $U$ for every $K_1$ that is a multiple of $3$ $\endgroup$
    – Asinomás
    Commented Aug 14, 2016 at 17:35
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    $\begingroup$ @Pinocchio - from your equation, $4K_1^2$ is divisible by $9$, right? Then, this means that $K_1^2$ itself must be divisible by $9$. (If you ask WHY, then you are in over your head - you should go back to basics). Then $K_1$ must be divisible by $3$. Put $K_1 = 3a$. That gives you Carry's solution. $\endgroup$
    – user325968
    Commented Aug 14, 2016 at 18:10

1 Answer 1


The solution set is $\{K_1=3a, U=3a+4a^2 \mid a\in \mathbb Z\}$

  • $\begingroup$ some more details for those dummys like me would be nice. Thanks for the help (can't accept yet). $\endgroup$ Commented Aug 14, 2016 at 17:46
  • $\begingroup$ @Carry on Smiling . You have a "typo". U needs to be 1/9 of what you wrote. $\endgroup$ Commented Aug 14, 2016 at 17:58
  • $\begingroup$ The edit by @mathguy is wrong. $\endgroup$ Commented Aug 14, 2016 at 18:26
  • $\begingroup$ @MarkL.Stone should be ok now, thanks for that. $\endgroup$
    – Asinomás
    Commented Aug 14, 2016 at 18:32

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