What a problem!

Geometrically, a triangular number can be arranged as a staircase. Two consecutive triangular numbers would make two staircase shapes, where one staircase has one more step than the other, and the smaller staircase can be flipped upside down and placed on top of the other to always form a square.

Therefore, it is true that two consecutive triangular numbers is always a square number.

A triangular number is defined as $\frac{n(n+1)}{2}$ where $n$ is an integer. Therefore, the sum of 2 consecutive triangular numbers is: $\frac{n(n+1)}{2}+\frac{(n+1)(n+2)}{2}$ $\frac{(n+1)(n)+(n+1)(n+2)}{2}$ $\frac{(n+1)(n+n+2)}{2}$ $\frac{2(n+1)(n+1)}{2}=(n+1)^2$

Therefore, the sum of two consecutive triangular numbers is always a perfect square. $~\beta_{\lceil \mid \rceil}$

The sum of consecutive triangular numbers is a square number, since,

$\begin{aligned} T_r + T_{r-1} & = \frac{1}{2} r \left(r+1 \right) + \frac{1}{2} \left(r-1 \right)r \\ & = \frac{1}{2} r \left[\left(r+1 \right)+ \left(r-1 \right)\right] \\ & = r^2 \end{aligned}$

Start by remembering that the formula for the $n^{th}$ triangular number is:

$T_n = \displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$

The formula for the next triangular number is thus:

$T_{n+1} = \displaystyle \sum_{k=1}^{n+1} k = \frac{(n+1)(n+2)}{2}$

Adding the two together:

$T_n + T_{n+1} = \frac{n(n+1)}{2} + \frac{(n+1)(n+2)}{2}$

$= (\frac{n+1}{2})(n+(n+2)) =(n+1)^2$

Which I think we can all agree is a perfect square .

An $n^{th}$ triangle number is calculated by $T = \frac{(n)(n+1)}{2}$ .

A square number for the $n^{th}$ term is very simple, $s = n^{2}$ .

We want to check whether two consecutive triangle number numbers ( $n$ and $n-1$ ) are equal to the $n^{th}$ term of a square number.

$n^{2} = \dfrac{(n-1)(n) + (n)(n+1)}{2}$ $= \dfrac{n^{2} - n + n^{2} + n}{2}$ $=\dfrac{2n^{2}}{2}$ $=n^{2}$

Observe the $n$ tringular number is bascially the sum of the first $n$ numbers, since it's constructed such there's there's $m$ points in the $m$ row.

Hence to prove the conjecture we shall generalize for two consequetive numbers, $n, n+1$ , and their sum $S$ : $\frac{n(n+1)}{2} + \frac{(n+1)(n+2)}{2} = S \\ n(n+1) + (n+1)(n+2) = 2S \\ (n+1)(2n+2) = 2n^{2} + 2n + 2n + 2 = 2n^{2} + 4n + 2 = 2(n^{2} + 2n + 1) = 2S \\ n^{2} + 2n + 1 = S \\ n^{2} + 2n + 1 = (n+1)^{2}$

Hence S will be a square number.

Without visualizing it: If you know squares are the sum of n odd numbers and triangles are the sum of all numbers up to n, when you add two consecutive triangle numbers you get double the smaller triangle number, which comes to the sum of n even numbers, and then you add n+1 for the next larger triangle, which means you can add 1 to all n even summands and change them from even to odd. And the +1 puts back the 1 on top and thus overall transforms this sum of even numbers to a sum of one more odd numbers and that makes it a square number.

Triangular numbers can be expressed by $\frac{n^2 + n}{2}$ . Therefore the sum of two consecutive numbers can be expressed as $\frac{n^2 + n}{2}$ + $\frac{(n+1)^2 + (n+1)}{2}$ , which simplifies to $\frac{n^2 +2n + 1}{1}$ = $\frac{(n+1)^2 }{1}$ , clearly a square number

Consider two triangular numbers they can be represented as $\frac{(n-1)\cdot n}{2}$ and $\frac{n\cdot (n+1)}{2}$ . The sum will be $\frac{2n^2}{2} = n^2$

The sum of two consecutive triangular numbers equals n(n + 1)/2 + (n + 1)(n + 2)/2 =( 2n^2 + 4n + 2)/2 = n + 1)^2. Ed Gray

The triangle having base with k Dots will have total Dots = K + (K-1) + (K-2) + ... + 1 which is equal to :-

K (K+1)/2. And a square with side R dots has R^2 Dots in total

Sum of 2 consecutive triangle =

K (K+1)/2 + (K+1)(K+2)/2

=( K+1)(K+1) = (K+1)^2 ie a square

I tried it with 10 and 15 and 10+15=25 which is a square root number of 5