Of course a triangle cannot be a square, but a triangular number can be a square number.

A triangular number is the sum of the first so many positive integers. For example, 10 is a triangular number because it equals 1+2+3+4. These numbers are called triangle numbers because you can form a triangle by having a row of one coin, two coin, three coins, etc. forming a triangle.

The smallest number that is both triangular and square is 1. The next smallest is 36. There are infinitely many numbers that are both triangular and square, and there’s even a formula for the *n*th number that is both a triangle and a square:

((17 + 12√2)^{n} + (17 – 12√2)^{n} – 2)/32

Source: American Mathematical Monthly, February 1962, page 169.

For more on triangle numbers and their generalizations, see Twelve Days of Christmas and Tetrahedral Numbers.

There is also a way to compute the square triangular numbers recursively discussed in the next post.

Here is a notebook evaluating the formula:

http://nbviewer.ipython.org/gist/certik/2945f1c08dc001314538

But, I bet you didn’t know that the parts of a triangle correspond to the sides of a rectangle.

http://youdothemathkthrucalculus.blogspot.com/2015/06/eureka-math-tips-for-parents-worst-sat.html

A triangle can’t be a square, but a triangle can have right angles at all three vertices — if the triangle lies on a sphere, and one vertex is a pole.

(actually, on second thought, none of the vertices need to be at a pole, it’s just easier to think about it that way — the triangle just needs to cover 1/4 of a hemisphere, oriented any way)

Should mention it’s reference at On-Line Encyclopedia of Integer Sequences. Good comments and more math trivia about the sequence…

https://oeis.org/A001110

I recently looked into which numbers are both triangular and hexagonal.

They are 1, 91, 8911, 873181, 85562821, …

A recursion is T(n)=99*T(n-1)-99*T(n-2)+T(n-3) and the direct formula

$latex 1/16+ \left( 3/16\,\sqrt {6}+{\frac {15}{32}} \right) \left( 49-20\,\sqrt {6} \right)^{n}+ \left( -3/16\,\sqrt {6}+{\frac {15}{32}} \right) \left( 49+20\,\sqrt {6} \right) ^{n}$.

Note that one of the terms is always small, giving an equivalent formula

$latex \left\lceil {\frac {1}{16}}+ \left( -3/16\,\sqrt {6}+{\frac {15}{32}} \right) \left( 49+20\,\sqrt {6} \right) ^{n}\right\rceil$.

http://oeis.org/A006244

n*(n+1)/2 = m^2

(2n+1)^2 – 8 m^2 = 1

Pell’s equation?

Yes, the problem can be transformed into a case of Pell’s equation. That’s where the solutions come from.