Let $T_n = \frac{1}{6}n(n+1)(n+2)$ denote the $n$th Tetrahedral number. The first several solutions to squares as sums of two Tetrahedral numbers are {T_n,T_m,a^2} 1 5 6\ 1 8 11\ 1 22 45\ 1 24 51\ 1 63 209\ 2 9 13\ 2 23 48\ 2 94 378\ 2 96 390\ 8 12 22\ 8 17 33\ 8 38 100\ 8 111 484\ 9 12 23\ 9 21 44\ 9 28 65\ 10 15 30\ 10 169 905\ 13 83 315\ 15 22 52\ 15 33 85\ 15 42 118\ 15 87 338\ 16 30 76\ 16 82 310\ 17 30 77\ 22 24 68\ 22 28 78\ 23 24 70\ 23 41 121\ 23 132 628\ 30 34 110\ 31 78 296\ 33 86 341\ 38 81 319\ 41 68 259\ 41 85 344\ 42 50 188\ 48 71 286\ 54 65 275\ 56 134 664\ 58 167 908\ 62 128 632\ 64 81 371\ 65 79 365\ 65 152 803\ 68 78 370\ 78 96 484\ 78 112 568\ 79 138 730\ 79 161 891\ 82 159 882\ 96 145 819\ 129 144 935\ Prove that there are infinitely many solutions.
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9$\begingroup$ No. ${}{}{}{}{}{}{}$ $\endgroup$– Will JagyCommented Jan 7, 2022 at 4:29
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$\begingroup$ If you admit a negative number $(m,n)$,we can show there are infinitely many solutions. $\endgroup$– TomitaCommented Jan 7, 2022 at 9:52
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1$\begingroup$ For what it's worth, 1, 6, 11, 45, 51, 209, 660099 are "Numbers $n$ such that $n^2 - 1$ is a tetrahedral number, see oeis.org/A216269 $\endgroup$– Gerry MyersonCommented Jan 7, 2022 at 15:59
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1$\begingroup$ There are two parametric solutions of $v^2=1/6n(n+1)(n+2)+1/6m(m+1)(m+2).$ $k$ is arbitrary integer. $(m,n,v)=( (6k+1)(2k+1), 6k^2+4k-1, (3k+1)(6k^2+4k+1) ),( (6k+5)(2k+1), 6k^2+8k+1, (3k+2)(6k^2+8k+3) ).$ $\endgroup$– TomitaCommented Jan 9, 2022 at 0:44
1 Answer
There are infinitely many integer solutions of equation $(1).$
$$y^2=\frac{n(n+1)(n+2)}{6}+\frac{m(m+1)(m+2)}{6}\tag{1}$$
Substitute $n = s - m$ to equation $(1)$ then we get
$$6y^2 = (3s+6)m^2+(-3s^2-6s)m+s^3+2s+3s^2$$
Let $s = 2t$ and $x = m - t$ then we get
$$3y^2 = (3t+3)x^2+t(t+2)(t+1)$$
Let $t = u^2-1$ then we get
$$y^2-u^2x^2 = \frac{(u^2-1)(u^2+1)u^2}{3}$$
Hence we get $(x,y)= \left(\Large{\frac{u^2+2}{3},\frac{2u^3+u}{3}}\right)$
To make $x$ and $y$ integers, let $u = 3k + 1$ then we get a parametric solution of euation $(1).$
$(m,n,y)=(\ (6k+1)(2k+1),\ 6k^2+4k-1,\ (3k+1)(6k^2+4k+1) \,)$
$k$ is arbitrary integer.
Similarly, $u = 3k + 2$ then we get another solution.
$(m,n,y)=(\ (2k+1)(6k+5),\ 6k^2+8k+1,\ (3k+2)(6k^2+8k+3) \,)$
Numerical example with $k\leqq 10.$
$(m,n,y)=(21, 9, 44),(65, 31, 231),(133, 65, 670),(225, 111, 1469),(341, 169, 2736),(481, 239, 4579),(645, 321, 7106),(833, 415, 10425),(1045, 521, 14644),(1281, 639, 19871),$
$(5, 1, 6), (33, 15, 85), (85, 41, 344), (161, 79, 891), (261, 129, 1834), (385, 191, 3281), (533, 265, 5340), (705, 351, 8119), (901, 449, 11726), (1121, 559, 16269), (1365, 681, 21856)$
