New years radical!!(in advance)

first, by simple manipulation, $\small{\small{\sqrt{\dfrac{(-a+b+\sqrt{a^2+b^2})(a-b+\sqrt{a^2+b^2})(a+b-\sqrt{a^2+b^2})}{4}\times 4(a+b+\sqrt{a^2+b^2})}}=4(a+b+\sqrt{a^2+b^2})}$ $\dfrac{1}{4}\small{\small{\sqrt{(a+b+\sqrt{a^2+b^2})(-a+b+\sqrt{a^2+b^2})(a-b+\sqrt{a^2+b^2})(a+b-\sqrt{a^2+b^2})}}}=a+b+\sqrt{a^2+b^2}$ now , $\quad a,b,\sqrt{a^2+b^2}\quad$ are sides of Pythagorean triangle. and $\dfrac{1}{4}\small{\small{\sqrt{(a+b+\sqrt{a^2+b^2})(-a+b+\sqrt{a^2+b^2})(a-b+\sqrt{a^2+b^2})(a+b-\sqrt{a^2+b^2})}}}$ is the area of a Pythagorean triangle by heron's. $$ the question is translated into " the right angle triangle with same area and perimetre" $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ re-write the equation as $a+b+\sqrt{a^2+b^2}=\dfrac{ab}{2}\rightarrow a^2 +b^2 =\dfrac{a^2b^2}{4}-ab(a+b)+a^2 +b^2 +2ab$ $\dfrac{a^2b^2}{4} + 2ab-ab^2-a^2b=0$ multiplying both sides by $\dfrac{4}{ab}$ $ab-4a-4b+8=0\longrightarrow (a-4)(b-4)=8$ 8 can be written as $8=2\times 4=1\times 8$ so $\begin{cases} a-4 =2\rightarrow a=6\\ b-4=4 \rightarrow b=8\\ a-4=1 \rightarrow a=5\\ b-4=8 \rightarrow b=12 \end{cases}$ $\therefore 6+8+5+12=\boxed{\color{#D61F06}{\large{3}}\color{#20A900}{\large{1}}}$