Two squared equals to three

$\overline { TWO } ^{ 2 }\quad =\quad \overline { THREE }$

For the above cryptarithm to be true, T can only be 1. But if W is 4 or larger, the above cryptarithm is also untrue. Therefore, W can either be 2 or 3. Let W = 2 and trying out O from 0, 3 to 9., there is no solution. For W =3, it is found that O = 8 gives the solution.

$138^{2} = 19044$

Therefore, T=1, W=3, H=9 and R=0 and

$T+W+H+R = 1+3+9+0 = \boxed{13}$

There is something wrong with the statement of the problem. It says that T,W,H and R are distinct integers. Not what E and O are.

Therefor, 140² = 19600 is also a right solution as is 130² = 16900. In both cases T,W, H and R are distinct integers. O and E aren't but that isn't required.

The cryptarithm is:

$138^2 = 19044$

& the value of $\mathsf{T} + \mathsf{W} + \mathsf{H} + \mathsf{R} = 1 + 3 + 9 + 0 = \boxed{13}$ .

$\text{ THREE has odd number of digits, and last digits of TWO and THREE }\\\text{are same. } ~~~\therefore~~~~T=1.~~~~~~~~~~~~~Let ~E_u ~ be ~ unit~ dgit.~~~~~~E_t~the~tenth. \\ \text{Since T=1, it recives no carry. So W=1, 2, or 3. We can see that the } \\\text{following conditions must hold good.} \\ \text{No two symbols can represent the same digit. So W=2,3 and O } \ne E_u. \\\implies\text{0,1,5,6 are eliminated from O and }E_u.~\therefore E_u ~ can~ not~ be~ 2,3,7, or~ 8.\\ \text{ Because no square of O will end in these.} ~E_u~ can~ only ~be~ 4,~ or~ 9. \\For ~~E_u \text{ to be 4, or 9..... O have to be 2 or 8, OR 3 or 7 . } \\Since~E_u=E_t, ~if~E_u=9, ~\text{ O= 3 or 7, an odd digit, so there is no}\\ \text{ carry or an even carry. Add to the carry the contribution of }W^2~ to ~ E_t.\\\text{This is always even. So this makes}~E_t ~an~ even~digit ! \implies E ~is~ only ~even.\\ \text{So only option remaining O=2 or 8 and W= 2 or 3. We can not have} \\\text{ W=O=2. So only options remaining are 128, 132, and 138.}\\ \text{So we see that }~138^2=19044.. ~~~~~~hits ~the ~target !!!!\\T+W+H+R=1+3+9+0=~~~\boxed{\color{#D61F06}{13}}$