We received good solutions from Chen Jinquan from The Chinese High School in Singapore and from Thomas from PS6 in New York.

Chen Jinquan explained that

From the applet, T₁₀ +T₁₀ is 10 x 11, and more generally T_n +T_n = n(n+1).

Hence, T_n =(n(n+1))/2,

which gives T₁₀ = 55,

T₆₀ = 1830,

T₁₀₀ = 5050,

T₂₅₀ = 31375,

T₂₀₄₅ = 2092035.

Consider 3655.

If 3655 is a triangular number, 7310 can be expressed as n(n+1).

By solving the quadratic equation or by estimating, we get n = 85, and hence 3655 = T₈₅

In general, a number x is a triangular number if and only if n(n+1)=2x is solvable for positive integers of n.

Consider 4851.

If it is a triangular number, 9702 can be expressed as n(n+1).

By solving the quadratic equation or by estimating, we get n = 98, and hence 4851 = T98

Consider 6214.

If it is a triangular number, 12428 can be expressed as n(n+1).

By solving the quadratic equation or by estimating, we see that there are no solutions in positive integers.

Hence 6214 is not a triangular number.

Using similar thinking leads to the conclusion that 7626 is T₁₂₃ and that 8656 is not a triangular number.

Thomas started off in a similar way but then used a different strategy for determining if numbers were triangle numbers:

T₁₀ +T₁₀ is a rectangle with sides 10 and 11 or area 110.

T₆₀ +T₆₀ is a rectangle with sides 60 and 61 or area 3660.

T₁₀₀ +T₁₀₀ is rectangle with sides 100 and 101 or area 10100.

So T₁₀ , T₆₀ , and T₁₀₀ are simply half the areas of the rectangles or 55, 1830, and 5050.

If you double a triangular number, you get a rectangular number. For the nth triangle number, the sides of the rectangle are n and n+1.

Therefore, the n^th triangular number can be written n(n+1)/2.

A strategy for finding triangle numbers is to put integers into the equation n(n+1)/2.

Using this T₂₅₀ = 31375

and T₂₀₄₅ = 2092035.

Since the n^th triangular number can be written n(n+1)/2, T is a triangle number if there is an integer n such that:

n² +n - 2T = 0

Solving this quadratic equation for n:

n = [ -1 + $\sqrt{(}1+8T)$ ] / 2

So T is triangular if $\sqrt{(}1+8T)$ is an integer.

[ -1 + $\sqrt{(}1+8x3655)$ ] / 2 = 85, so 3655 is the 85^th triangle number.

$\sqrt{(}1+8T)$ gives integers for 4851 and 7626, so they are triangle numbers.

6214 and 8656 are not.

Well done to you both.