Are there infinitely many squares in the set $$\{\sum_{j=1}^m j^2: m\in\mathbb{N}\} ?$$
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asked Apr 13, 2019 at 17:01
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7$\begingroup$ No, only for $m=1$ and $m=24$. math.stackexchange.com/questions/36702/… $\endgroup$– literature-searcherCommented Apr 13, 2019 at 17:53
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6$\begingroup$ This is the well-known square pyramid problem whose only solutions are $m=1$ and $m=24$, although the proof is not immediate. $\endgroup$– Henri CohenCommented Apr 13, 2019 at 17:53
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$\begingroup$ See also: The sum of the first $n$ squares is a square: a system of two Pell-type-equations (and other questions linked there). $\endgroup$– Martin SleziakCommented Apr 13, 2019 at 23:31
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1 Answer
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We have $\sum_{j=1}^m j^2 = \frac{m(m+1)(2m+1)}6$. Hence, the question reduces to finding integral points on the elliptic curve: $$y^2 = \frac{x(x+1)(2x+1)}6.$$ Turning it to Weierstrass equation with integer coefficients, we get: $$(72y)^2 = (12x)^3 + 18 (12x)^2 + 72 (12x).$$
For many curves, integral points can be readily found by SageMath, and in this case we are lucky:
sage: EllipticCurve([0,18,0,72,0]).integral_points()
[(-12 : 0 : 1), (-9 : 9 : 1), (-8 : 8 : 1), (-6 : 0 : 1), (0 : 0 : 1), (6 : 36 : 1), (12 : 72 : 1), (288 : 5040 : 1)]
So, the only solutions are $m=\frac{0}{12}=0$, $m=\frac{12}{12}=1$, and $m=\frac{288}{12}=24$.
answered Apr 13, 2019 at 20:42