## When the sum of two Triangular numbers is a perfect cube

Since   $T_n + T_{n-1} = n^2$

we may write

$T_{n^3} + T_{n^3 - 1}$

$n^3(n^3 + 1)/2 + (n^3 - 1)(n^3)/2 = (n^2)^3$

we find an infinite number of such triangular pairs,

$T_1 + T_0 = 1 + 0 = 1^3$
$T_{2^3} + T_{2^3 - 1} = 36 + 28 = 64 = 4^3$
$T_{3^3} + T_{3^3 - 1} = 378 + 351 = 729 = 9^3$
$T_{4^3} + T_{4^3 - 1} = 2080 + 2016 = 4096 = 16^3$
$T_{5^3} + T_{5^3 - 1} = 7875 + 7750 = 15625 = 25^3$
$T_{6^3} + T_{6^3 - 1} = 23436 + 23220 = 46656 = 36^3$
$T_{7^3} + T_{7^3 - 1} = 58996 + 58653 = 117649 = 49^3$
$T_{8^3} + T_{8^3 - 1} = 131328 + 130816 = 262144 = 64^3$
$T_{9^3} + T_{9^3 - 1} = 266085 + 265356 = 531441 = 81^3$
$T_{10^3} + T_{10^3 - 1} = 500500 + 499500 = 1000000 = 100^3$
…………
…………..

but not EVERY triangular pair can be found in this manner

Other solutions include:

Solutions of the form    $T_n + T_{2n}$ :

$T_3 + T_6 = 6 + 21 = 27 = 3^3$
$T_9 + T_{18} = 45 + 171 = 216 = 6^3$
$T_{48} + T_{96} = 1176 + 4656 = 5832 = 18^3$

Solutions of the form    $T_n + T_{n + k}$ :

$T_3 + T_6 = 27 = 3^3 = (3\times 1)^3$

$T_9 + T_{18} = 216 = 6^3 = (3\times 2)^3$
$T_3 + T_{20} = 216 = 6^3 = (3\times 2)^3$

$T_{26} + T_{27} = 729 = 9^3 = (3\times 3)^3$

$T_{68} + T_{83} = 5832 = 18^3 = (3\times 6)^3$
$T_{48} + T_{96} = 5832 = 18^3 = (3\times 6)^3$

$T_{93} + T_{137} = 13824 = 24^3 = (3\times 8)^3$

$T_{134} + T_{189} = 27000 = 30^3 = (3\times 10)^3$
$T_{29} + T_{230} = 27000 = 30^3 = (3\times 10)^3$

$T_{147} + T_{267} = 46656 = 36^3 = (3\times 12)^3$
$T_{125} + T_{278} = 46656 = 36^3 = (3\times 12)^3$

Note that,
$125 = 5^3$   and
$278 = 125 + 153 = 5^3 + T_{17}$

$T_{42} + T_{302} = 46656 = 36^3 = (3\times 12)^3$

$T_{129} + T_{362} = 74088 = 42^3 = (3\times 14)^3$

$T_{150} + T_{399} = 91125 = 45^3 = (3\times 15)^3$
$T_{185} + T_{384} = 91125 = 45^3 = (3\times 15)^3$

$T_{326} + T_{338} = 110592 = 48^3 = (3\times 16)^3$
$T_{156} + T_{443} = 110592 = 48^3 = (3\times 16)^3$

$T_{191} + T_{527} = 157464 = 54^3 = (3\times 18)^3$

$T_{140} + T_{852} = 373248 = 72^3 = (3\times 24)^3$

$T_{644} + T_{654} = 421875 = 75^3 = (3\times 25)^3$
$T_{450} + T_{800} = 421875 = 75^3 = (3\times 25)^3$