<s>[INST] Triangle numbers problem 
Find the smallest three-digit triangular number that can be
  represented both as the sum of three different triangular
  numbers, and as the sum of two different triangular numbers.

So the answer is $120$, but I found it by trial and error. What is the clever 'mathematical' way of answering this question?
 [/INST] I went through it. With $T(x) = (x^2 + x)/2,$ we start with
$$  T(u) = T(v)+T(w) = T(x) + T(y) + T(z).  $$
Multiply by $8.$
Adding $1$ to all three positions and considering the first equality we get
$$ (2u+1)^2 + 1 = (2v+1)^2 + (2w+1)^2.  $$
This is always solvable, except that you want all three letters positive and $v \neq w.$ The short version is that we must have $ (2u+1)^2 + 1$ divisible by two distinct primes $\equiv 1 \pmod 4.$ For example, if $2u+1 = 13,$ we have $170,$ which can indeed be written as $11^2 + 7^2.$
Second thing to check: instead of $1,$ add $2$ to all positions and equate the first position and the third. We have
$$ (2u+1)^2 + 2 = (2x+1)^2 + (2y+1)^2 + (2z+1)^2.  $$
Once again this is always possible, except you want $x,y,z$ positive and distinct. So we strike out as
$$ 171 = 11^2 + 5^2 + 5^2 = 9^2 + 9^2 + 3^2.  $$
When $2u+1 = 21,$ we do get $442 = 17 \cdot 13 \cdot 2,$ and indeed
$$ 21^2 + 1 = 442 =  9^2 + 19^2.  $$ Then
$$  443 = 15^2 + 13^2 + 7^2.  $$
So that works. And the first occasion is actually 
$$  \color{magenta}{55 = 10+45 = 6+21+28}.   $$
I see, they want at least $100.$
As $u$ gets larger, the ability to solve this with the full prohibitions should become the norm rather than the exception.
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