Show number of integer solutions of k^2 +k=2t^2 is infinite,k&t are integers?

Answer 1

Multiply both sides by 4 and add 1 to each side.
We get 4k²+4k+1 = (2k+1)² = 8t²+1.
Now let u = 2k +1
and u²-8t²=1.
This is a Pell equation and, by the theory of such
equations, it has infinitely many solutions. In fact,
the smallest solution is u = 3, t = 1,
so all integer solutions are given by
u + √t = ±(3+√8)^n,
where n is any integer.
Let's find some solutions, using the + sign and
positive n.
n = 1: (3,1),
n = 2: (17,6)
n = 3: (99,35)
n = 4: (577,204).
Hope that helps!

Answer 2

This is equivalent to the solutions of
k (k + 1) / 2 = t²
or the triangular numbers which are also square.
The triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91 ...
Observe it to be true at k = 8, t = 6 and at k = 1, t = 1.

There is a good recursion-type approach to this. I can't remember it, though. If I do figure it out, I will tell you.

The solution I found produces each successive solution, but also solves the problem directly, without Pell's equation.