Was googling your problem and came across this:

It turns out that there are only 2 primes that

satisfy 8p+120 = a triangular number.

Clearly, p = 2 works:

16+120 = 136 = 16*17/2.

We will show that the only other value of p

that works is p = 47.

Thus the sum of all the required primes is 49.

We need the following result:

If n is a triangular number then 8n+1 is a square.

So we can write the condition of the problem as

64p + 961 = x^2

or

(x+31)(x-31) = 64p.

Suppose p = 2

Since x > 31 we have x -31 = 2

x = 33, 128 + 961 = 1089 = 33^2.

So let's assume p is an odd prime

Then, because p is prime, we must have

one of the following:

a. x-31 = 32, x+31 = 2p

Then x = 63, 63+31 = 94

94*32 + 961 = 3969, a square.

So p = 47. 8p+120 = 496 = 31*32/2.

b. x-31 = 16, x+31 = 4p

x = 47. 47+31 = 78, but

78 is not of the form 4p.

c). x-31 = 8. x+31 = 8p

x = 39, 39+31 = 70,

but 70 is not of the form 8p.

Similarly, the remaining 2 cases are impossible.

The answer to the problem is therefore 49.