I am working on solving the following Diophantine equation:
$$k^2(k+1)=m(3m-1)$$
And so far I solved, using WolframAlpha, the following solutions:
$$(k,m)=(-1,0);(0,0);(1,1);(4,-5);(6,-9)$$
Is there a way to prove that those are the only ones?
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Hint: This is an elliptic curve given by $3y^2-y=x^3+x^2$. Its integer points can be computed, see for example the following post (we might transform to short Weierstrass form $y^2=x^3+ax+b$ if necessary).
How to compute rational or integer points on elliptic curves
Dietrich Burde
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Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Jan 17 '20 at 18:40
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(k,m)=(-2/3),(-1/9) & (k,m)=(-2/3),(4/9)
The above are also the solutions to the equation asked by "OP".
Mathew
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$k^2(k+1)=m(3m-1)$
⇒$3m^2-m-k^2(k+1)=0$
⇒ $m=\frac{1±\sqrt {\Delta}}{6}$
The discriminant $\Delta$ is:
$\Delta=1+12k^2(k+1)≥ 0$
Only $k=-1, 0, 1, 4, 6 $ give perfect square for $\Delta$ .
– sirous Jan 18 '20 at 10:55