Assume that the sides are x2+y2, 2xy, and x2-y2 and you'll get 6,8,10 and 5,12,13

Sorry, it's only for right-angled triangle.

Heron' formula gives $S=\frac{1}{2}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$. Since $a+b+c=S$ then we have $4(a+b+c)=(a+b-c)(b+c-a)(c+a-b)$, or $4( a+b-c+b+c-a +c+a-b)=(a+b-c)(b+c-a)(c+a-b)$. Let $a+b-c=p, b+c-a=q$ and $c+a-b=r$, then we have to solve the Diophantine equation $4p +4q+4r=pqr$. Without loss of generality, we may assume $p\geqslant r \geqslant q$, then $pqr \leqslant 12p$ or $qr \leqslant 12$. The rest is just considering the cases when $qr=12, qr=11 $ etc, and I'm too lazy to do that.

enndb0x wrote: Heron' formula gives $S=\frac{1}{2}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$. Since $a+b+c=S$ then we have $4(a+b+c)=(a+b-c)(b+c-a)(c+a-b)$, or $4( a+b-c+b+c-a +c+a-b)=(a+b-c)(b+c-a)(c+a-b)$. Let $a+b-c=p, b+c-a=q$ and $c+a-b=r$, then we have to solve the Diophantine equation $4p +4q+4r=pqr$. Without loss of generality, we may assume $p\geqslant r \geqslant q$, then $pqr \leqslant 12p$ or $qr \leqslant 12$. The rest is just considering the cases when $qr=12, qr=11 $ etc, and I'm too lazy to do that. I was started at that way, but we have $16(a+b+c)=(a+b-c)(b+c-a)(c+a-b)$ because you made a mistake in the theorem ( now I know I'm not the only one who does). At your way we have $pqr=16(p+q+r).$ and that will be a lot of cases, because we will get $qr<48$ and I don't want look to so many cases. It is easy that $qr$ isn't prime and bigger than $16$, but yet then e have a lot of work. Anyone a shorter solution, because the solution we have is also with $8$ cases.

You can find the end of the solution using Heron's formula here: http://jwilson.coe.uga.edu/emt725/perfect/sol.html It confirmes the five solutions found by Maharjun above, and gives a reference to Mathematics Magazine no. 1, 1969, p. 47.

Titanium wrote: You can find the end of the solution using Heron's formula here: http://jwilson.coe.uga.edu/emt725/perfect/sol.html It confirmes the five solutions found by Maharjun above, and gives a reference to Mathematics Magazine no. 1, 1969, p. 47. Am I so wrong?? Is Heron's formula not dividing by $4?$

I just want to put a slight filler in my post by heros formula, let $p = a + b+c$ $16(p) = (p - 2a)(p - 2b)(p - 2c)$ since one of $(p - 2a),(p - 2b),(p - 2c)$ must be even hence $p$ is even $\implies (p - 2a),(p - 2b),(p - 2c)$ are all even $\implies s-a, s-b, s-c$ must also necessarily be integers [i believe this was not clear in my previous post] [the quantities $x,y,z$ are basically these] a second thing which i think may be worth noting is that if any one of $(s-a),(s-b),(s-c)$ is $2$ it must be a right triangle whether or not the other two are integers or not because that would mean $s-a = r$ and this implies triangle right angled at $A$

Like in the previous solutions, we use Heron formula to obtain the equation $16(a+b+c)=(-a+b+c)(a-b+c)(a+b-c)$ and the substitution $-a+b+c=x$ and cyclic ones to reduce everything to $16(x+y+z)=xyz$. Now we distinguish in cases, according to if $x, y, z$ are: - all odd: then there is obviously no solution since the LHS is a multiple of $16$ while RHS is not. - one even and two odd: dividing by cases according to the parity of $a, b, c$, we find that $x, y, z$ must be all even or all odd, so there is no solution. - two even and one odd: as before, we have no solutions. - all even: set $x=2t, y=2u, z=2v$ to get the equation $tuv=4(t+u+v)$, so $v=\frac {4(t+u)} {tu-4}$; this implies that $\min(t, u) \leq 8$; for fixed values of $u$ it is easy to find all integer values of the fraction using the Euclidean algorithm, so trying the possible values for $u$ we find the solutions $(a, b, c)=(5, 12, 13), (6, 8, 10), (6, 25, 29), (7, 15, 20), (9, 10, 17)$.

Hello, is there any bibliographical resource about that problem? Except from that particular website, i can't find any bibliographical notes about it. What should i do?

I have used Ravi substitution it is much easier now to solve it

Smita wrote: I have used Ravi substitution it is much easier now to solve it I'd like to see how you did it!