<s>[INST] The Jacobi-Madden equation $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ and disguised Pythagorean triples I. The Jacobi-Madden equation,
$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4$$
is equivalent to a disguised Pythagorean triple,
$$(a^2+ab+b^2)^2+(c^2+cd+d^2)^2 = \big((a+b)^2+(a+b)(c+d)+(c+d)^2\big)^2$$

II. A special case of the Descartes' circle theorem,
$$2(a^4+b^4+c^4+d^4)=(a^2+b^2+c^2+d^2)^2$$
and Euler showed this is just,
$$(2ab)^2+(2cd)^2 = (a^2+b^2-c^2-d^2)^2$$

III. The Fermat quartic,
$$a^4+b^4 = c^4$$
becomes,
$$(a b - a c + b c + c^2)^2 + (a b + a c - b c + c^2)^2 = (a^ 2 + b^2 + c^2)^2$$

IV. The Pell equation,
$$x^2-2y^2 = -1$$
is also,
$$x^2 + (y^2 - 1)^2 = y^4$$
as well as,
$$\Big(\frac{x-1}{2}\Big)^2+\Big(\frac{x+1}{2}\Big)^2 = y^2$$
with the latter showing there are infinitely many triples where the legs differ by just $1$.

Q: Are there any other examples of simple quadratic or quartic equations that can be expressed as a Pythagorean triple?

 [/INST] Not sure if this is what you had in mind, but suppose we have two distinct triangular numbers whose product is a square. Say $[u(u+1)/2][v(v+1)/2] = t^2$, where  $u > v > 0$. Define $a = 4t$, $b_1 = u - v$, $b_2 = u + v + 1$, $c = 2uv + u + v$. Then a routine calculation gives
$$a^2 + b_1^2 = c^2,\quad a^2 + b_2^2 = (c + 1)^2.$$
Thus we get two Pythagorean triangles with a common side and hypotenuses differing by $1$. For instance $u = 8$, $v = 1$ gives $(24, 7, 25)$ and $(24, 10, 26)$. Alternatively you can start with two such triangles and reverse the procedure, as is easily proved.
</s>