Happy Triangles

The best solution to this problem is to refer to an equation known as Heron's Formula. This is an equation for the area of a triangle that requires only the lengths of its sides. It is as follows:

For a triangle with sides $a$ , $b$ , & $c$ :

$\large A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \Large\frac{a+b+c}{2}$

Our question asks for triangles with consecutive side lengths. We'll let $a = x-1$ , $b = x$ , and $c = x+1$ . This means that our solution will be of the form $3x$ . First we'll plug in out numbers into our formula for $s$ .

$s = \Large\frac{x-1+x+x+1}{2} = \frac{3}{2} x$

Now we'll plug in our values of $s$ , $a$ , $b$ , & $c$ into our formula for the area.

$\large A = \sqrt{\frac{3}{2} x(\frac{3}{2} x-x+1)( \frac{3}{2} x-x)( \frac{3}{2} x-x-1)}$

$\large A = \sqrt{\frac{3}{2} x(\frac{1}{2} x+1)( \frac{1}{2} x)( \frac{1}{2} x-1)}$

$\large A = \frac{x}{2} \sqrt{3(\frac{1}{2} x+1)( \frac{1}{2} x-1)}$

$\large A = \frac{x}{2} \sqrt{3(\frac{1}{4} x^2-1)}$

$\large A = \frac{x}{4} \sqrt{3(x^2-4)}$

For $A$ to be an integer, we need two things to happen:

$1$ . $3(x^2-4)$ has to be a square number

$2$ . We need to check if $4\mid x\sqrt{3(x^2-4)}$

Let's begin by finding numbers that satisfy $3(x^2-4) = n^2$

Observe that $x$ cannot be divisible by 3 because we need $3\mid x^2-4$ to be true.

$(3n)^2 \equiv 0$ (mod $3$ ) $\Longrightarrow (3n)^2-4 \equiv 2$ (mod $3$ )

Also not that $x$ cannot be odd. If $x$ was odd, the $3(x^2-4)$ would have to be an odd square, and $x\sqrt{3(x^2-4)}$ would be odd as well, meaning that $4\nmid x\sqrt{3(x^2-4)}$

This leaves us to check only numbers of the form 6n-2 and 6n+2. This narrows down our search a bit.

$3((6n-2)^2-4)=3(36n^2-24n+4-4)=6^2(3n^2-2n)$

$3((6n+2)^2-4)=3(36n^2+24n+4-4)=6^2(3n^2+2n)$

So either $3n^2-2n$ or $3n^2+2n$ must be a perfect square.

From here it's pretty simple to check values of $n$ to see what works and what doesn't. Letting $n=1$ in our 1st equation yields $x=4$ as a solution, and letting $n=2)$ yields $x=14$ , and our next solution doesn't come until $n=9$ where $3\times9^2-2\times9 = 225 = 15^2$

$x=6(9)-2=52 \Longrightarrow 3x = \boxed{156}$