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During solving another equation, I found equation $1+12k^2(k+1)=t^2$ has limited solution in $\mathbb Z$:

$(k, t)=(0,±1), (1, ±5), (-1, ±1), (4,±31), (6,±55)$

I found this to be true by Python up to $k=10^6$. Prove or disprove this analytically.

J. W. Tanner
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sirous
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    Now we have a circle of posts regarding this question. It started with this post, then went here and here... – Dietrich Burde Jan 15 '20 at 15:21
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    ... coming back now at the present question to the first post asking again for $1+12k^2(k+1)=t^2$, see one answer at this original post. Now the answer by Robert closes the circle. Perhaps this could be joined somehow and superflous posts deleted or combined. – Dietrich Burde Jan 15 '20 at 15:24

1 Answers1

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This is an elliptic curve. According to a theorem of Siegel, it has only finitely many integer points.

EDIT: The Weierstrass form of this curve is $$x^3 + y^2 - 48 x - 272 = 0 $$ where $x = -12 k - 4$ and $y = 12 t$. Note that integer $(k,t)$ imply integer $(x,y)$ but not vice versa. Sage gives me the integer points $(x,y)$:

x, y = var('x,y')
E = EllipticCurve(x^3 - 48*x + 272 == y^2)
E.integral_points(both_signs=True)

[(-8 : -12 : 1),
 (-8 : 12 : 1),
 (-4 : -20 : 1),
 (-4 : 20 : 1),
 (1 : -15 : 1), 
 (1 : 15 : 1),
 (4 : -12 : 1),
 (4 : 12 : 1),
 (8 : -20 : 1), 
 (8 : 20 : 1), 
 (16 : -60 : 1), 
 (16 : 60 : 1), 
 (52 : -372 : 1), 
 (52 : 372 : 1), 
 (76 : -660 : 1), 
 (76 : 660 : 1), 
 (1721 : -71395 : 1), 
 (1721 : 71395 : 1)]

Looking at the corresponding $(k,t)$ pairs, the only ones in integers are those you found.

Robert Israel
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