Let's find two sets of natural numbers, preferably small, such that when used as the side lengths of a triangle, all four factors in Heron's formula for the area of the triangle turn out to be perfect squares. One of these sets should form an isosceles triangle, and the other should form a triangle with all different side lengths.
Using the conventional notation, the requirement is:

$$
s=k^{2}, \quad s-a=l^{2}, \quad s-b=m^{2}, \quad s-c=n^{2}
$$

where $k, l, m, n$ are natural numbers. By equating the sum of the last three equations with the first one,

$$
(s-a)+(s-b)+(s-c)=3 s-(a+b+c)=3 s-2 s=s
$$

it must be true for the sought natural numbers that

$$
l^{2}+m^{2}+n^{2}=k^{2}
$$

This equation provides a suitable triplet of numbers:

$$
a=s-l^{2}=k^{2}-l^{2}, \quad b=k^{2}-m^{2}, \quad c=k^{2}-n^{2}
$$

indeed, for these, the triangle inequalities must necessarily hold, for example, if

$$
0 < a < b + c
$$

We will show that by choosing $l$ as any natural number and $m$ as any even number, we get a solution to (1). Indeed, with $l^{2}+m^{2}=M$,

$$
k^{2}-n^{2}=(k-n)(k+n)=M
$$

and here $M$ is always factorizable into a pair of natural numbers $u, v$ (with $u < v$), because in (3) the numerators of the fractions are even numbers; if $l$ is an even number, then $M$ is divisible by 4, because both terms are divisible by it, so $u=2$ and $v=M / u=M / 2(>2)$ are both even, and $k$ and $n$ are again natural numbers.

In any case, an isosceles triangle is obtained by choosing $l=m$. For example, let $l=m=2$, so $M=8$; with the choice $u=2$ and $v=4$ in (3), and then in (2), we get $k=3, n=1, a=b=5, c=8, s=9$ and $t=k l m n=12$.

The choice $l \neq m$ does not guarantee that the sides of the triangle will be different - for example, $l=1, m=2$ leads to the previous example - but after a few trials, we can find a solution that satisfies the second requirement: starting from the pair $l=3, m=2$, we get $k=7, n=6, a=40, b=45, c=13 ; s=49, t=252$.

András Koren (Budapest, I. István g. II. o. t.)

Remarks. 1. We can also find a solution to (1) by taking a random Pythagorean triplet - for example, 3, 4, 5 - and finding another one where the hypotenuse of the first triplet plays the role of one of the legs, continuing the example with 5, 12, 13. Thus, $k=13, l=3, m=4, n=12 ; a=160, b=153, c=25$. However, a smaller triangle side triplet can also be found.

László Mérő (Budapest, Berzsenyi D. g. II. o. t.)

2. In several papers, it is readable that we should look for an identity similar to the one giving Pythagorean triplets, $\left(u^{2}-v^{2}\right)^{2}+(2 u v)^{2}=\left(u^{2}+v^{2}\right)^{2}$, from which multiple solutions to (1) can be generated (since such an identity is not found in commonly available collections). Some have also shared such identities.

For every even $m$ number, there exists a pair of integers $l, n$ such that $2 l n=m^{2}$. Then the left side of (1) is $l^{2}+2 l n+n^{2}=(l+n)^{2}$, so $k=l+m^{2} / 2 l$. (For example, $m=2, l=1, n=2, k=3$.) Eliminating the fractions, the identity $\left(2 l^{2}\right)^{2}+(2 l m)^{2}+\left(m^{2}\right)^{2}=\left(2 l^{2}+m^{2}\right)^{2}$ is obtained, which provides a solution for any positive integer pair of $l, n$ parameters (but does not provide all solutions; Péter Rajczy, Bp., Eötvös Gymn.).

Another 2-parameter solution: $(2 u v)^{2}+(4 u v)^{2}+\left(u^{2}-5 v^{2}\right)^{2}=\left(u^{2}+5 v^{2}\right)^{2}$ (Péter Moson, Bp., Fazekas Gymn.).

A 1-parameter solution: $u^{2}+(u+1)^{2}+[u(u+1)]^{2}=[u(u+1)+1]^{2}$ (László Takács, Sopron, Széchenyi Gymn.). From this, for $u=3$, the already seen 3, 4, 12, 13 solution is obtained.

See also the 1085. exercise set in this issue.