Special triangles

Let positive integers $a,b$ be the legs of the right triangle and $c = \sqrt{a^{2} + b^{2}}$ be the hypotenuse. Then we require that

$a + b + c = \dfrac{ab}{2} \Longrightarrow ab - 2(a + b) = 2c \Longrightarrow (ab)^{2} - 4ab(a + b) + 4(a + b)^{2} = 4c^{2} \Longrightarrow$

$ab(ab - 4a - 4b) + 4(a^{2} + 2ab + b^{2}) = 4c^{2} = 4(a^{2} + b^{2}) \Longrightarrow ab(ab - 4a - 4b + 8) = 0 \Longrightarrow$

$ab - 4a - 4b + 8 = 0 \Longrightarrow (a - 4)(b - 4) = 8$ .

Now $8$ can be factored, without concern about ordering, as $(-8)*(-1), (-4)*(-2), 4*2$ and $8*1$ . Assigning the negative factors to $a - 4$ and $b - 4$ yields at least one negative value for $a,b$ , so the only valid assignments involve the $4*2$ and $8*1$ factorings. Because of the symmetry between $a,b$ in the product $(a - 4)(b - 4)$ we can, without loss of generality, take $a \le b$ , in which case we can either have

$a - 4 = 2, b - 4 = 4 \Longrightarrow (a,b,c) = (6,8,10)$ or

$a - 4 = 1, b - 4 = 8 \Longrightarrow (a,b,c) = (5,12,13)$ .

So there a total of $\boxed{2}$ suitable integer-sided right triangles.

$We~know~that~pythagorean~triples~are~generated~by~integers~m>n,\\ where~the ~triples~are~(m^2-n^2, 2mn, m^2+n^2).\\ \therefore~we~ are~conditioned~by~area=perimeter.\\ \implies~\frac 1 2*(m^2-n^2)*2mn=m^2-n^2+ 2mn+ m^2+n^2=2m(m+n).\\ However~m\neq 0,~and~m+n\neq 0. ~So~dividing~both~sides~with~their~product,\\ (m-n)n=2,~~\implies~it~is~a~quadratic~in ~n.\\ \therefore~\color{#E81990}{n=\dfrac{m\pm \sqrt{m^2-8}} 2.}\\ So~m>2.~~Let~m=3, ~we~get~n=1~0r~n=2.\\ As~we~see~below~m~can~not~take~any~other~value.\\ Only~TWO~solutions~with~(m,n)=(1,3)~and~ (2,3)~ ~are~~(6,8,10)~and~(5,12,13).\\ But~for~n~to~be~an~integer,~m^2-8=k^2,~k~some~integer.\\ \implies (m-k)(m+k)=2*4,~or~1*8.\\ So~m-k=2,~and~m+k=4.~\therefore~k=3.~~~We~ have~ seen~ this.\\ \or~~~m-k=1~and~m+k=9.~\therefore~k=5.\\ For~m=5,~~\sqrt{25-8}~is~ not~ an~ integer.\\$

Given: $a + b + c = ab/2$ $a^2 + b^2 = c^2 \\ (a + b)^2 = c^2 + 2ab \\ 2ab = (a + b + c)(a + b - c)$

$4a + 4b + 4c = 2ab = (a + b + c)(a + b - c) \\ 4(a + b + c) = (a + b + c)(a + b - c) \\ a + b - c = 4 \\ \implies a + b = 4 + c$

$a + b + c = ab/2 \\ \implies ab = 8 + 4c$

$\implies x^2 - (4+c)x + 8+4c = 0 \\ \left( x - \cfrac{4+c}{2}\right )^2 = \cfrac{16 + 8c + c^2 - 32 - 16c}{4} = \cfrac{c^2 - 8c - 16}{4} = \cfrac{(c - 4)^2 - 32}{4} \\ x = \cfrac{4+c \pm \sqrt{(c - 4)^2 - 32}}{2}$

if c is odd: then, c + 4 will turn out odd and hence, $(c - 4)^2 - 32$ should be equal to 1 mod 4. so, $(c-4)^2 - 32 \equiv 1 \mod 4 \equiv (4n + 1) \mod 4 \\ (c - 4)^2 = 4n + 33$

since $33 \mod 4 \equiv 1 \implies 4n \mod 4 = 0 \implies n = 12 \implies c = 13$ since if n = 4, c = 11 which is not a possible hypotenuse of a right triangle with integer sides.

if c is even: then, c + 4 will turn out even and hence, $(c - 4)^2 - 32$ should be equal to 0 mod 4. so, $(c-4)^2 - 32 \equiv 0 \mod 4 \equiv (4n) \mod 4 \\ (c - 4)^2 = 4n + 32 = 4(n + 8) \\ 4n \mod 4 \equiv 0 \implies n = 1 \implies c = 10 .$

since there are 2 (only) possible values of c, therefore, there are (only) $\boxed{2 \text{right triangles}}$ with integer sides such that, their perimeter and area are equal.

$\text{ I know, my solution have holes. Can you help me clear out that holes? i.e. How can you prove that } \\(c - 4)^2 = 4n + 33 \ and \ (c - 4)^2 = 4n + 32 \\ \text{ have only one feasible value of n such that, it will give a integer value c and it can be a possible hypotenuse of a right triangle.?}$