Find all integers $m$ and $n$ such that $$m^3-m=3n(n+1)$$
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$m^3-m$ is divisible by $3, \forall m\geq 1$, and, $3n(n+1)$ is also divisible by $3$, could be somewhere you can go from that. – Plus Twenty Jun 09 '19 at 05:50
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See https://www.quora.com/What-are-the-integer-solutions-to-m-3-m-3n-n-1 – nmasanta Jun 09 '19 at 05:58
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1Notice that $ m^3 - m $ is the multiplication of three consecutive numbers: $ m^3 - m = (m-1) m (m+1) $. Any triplet of consecutive numbers will always be a multiple of 3. – Duns Jun 09 '19 at 06:19
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1This is ${m+1\choose3}={n+1\choose2}$ which I suspect has been discussed here (or maybe at mathoverflow) beofre. – Gerry Myerson Jun 09 '19 at 06:39
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Any thoughts on my answer, 679770? – Gerry Myerson Jun 11 '19 at 05:11
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Are you still here, 679770? – Gerry Myerson Jun 12 '19 at 13:13
1 Answers
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Numbers that are both triangular and tetrahedral are tabulated at http://oeis.org/A027568 and they are $0, 1, 10, 120, 1540, 7140$. $m=4,n=4$ corresponds to $10$, $m=9,n=15$ gives $120$, the others, see below. There are references and links at the OEIS page.
$$1540={22\choose3}={56\choose2}$$
$$7140={36\choose3}={120\choose2}$$
In case the connection to the original question is not clear, ${m+1\choose3}={n+1\choose2}$ is $(m+1)m(m-1)/6=n(n+1)/2$, which is $m^3-m=3n(n+1)$.
Gerry Myerson
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See also https://math.stackexchange.com/questions/2214853/how-often-can-a-number-occur-in-pascals-triangle and https://math.stackexchange.com/questions/2052569/integral-solutions-to-1-times22-times3-cdotsm-timesm1-n-timesn1 – Gerry Myerson Jun 09 '19 at 07:15
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And https://math.stackexchange.com/questions/2294675/when-do-different-binomial-coefficients-coincide and https://math.stackexchange.com/questions/85442/are-there-surprisingly-identical-binomial-coefficients – Gerry Myerson Jun 09 '19 at 07:41