fourth powers as sums of squares Is it possible to have a fourth power that is the sum of two squares in four different ways, e.g., $w^4 = a^2 + b^2 = c^2 + d^2 = e^2 + f^2 = g^2 + h^2$ with the added restriction that $e = a+c$ and $g = a-c$ ?  What is the lowest example, or why is it impossible?
 A: Since $e=a+c$ and $g=a-c$, then the OP wishes to solve the system,
$$a^2 + b^2= c^2 + d^2\tag1$$ 
$$a^2 + b^2= (a+c)^2 + f^2\tag2$$
$$a^2 + b^2= (a-c)^2 + h^2\tag3$$ 
$$a^2 + b^2= w^4\tag4$$
It turns out that, using an elliptic curve, there is an infinite number of primitive integer solutions to the first $3$ conditions. However, it is highly doubtful the $4$th can be satisfied as well.
The first was completely solved by Brahmagupta as,
$$a,\;b,\;c,\;d = p r + q s,\; p s - q r,\; p r - q s,\; p s + q r$$
The second and third become,
$$-(3 p^2 - q^2) r^2 + (p^2 + q^2) s^2 = f^2$$
$$(p^2 + q^2) r^2 + (p^2 - 3q^2) s^2 = h^2$$
For example, let $p=1,\;q=2$, then,
$$r^2+5s^2=f^2\tag5$$
$$5r^2-11s^2=h^2\tag6$$
with initial rational point $r=178,\; s=19$. Two quadratic polynomials to be made a square and that has a rational point can be transformed into an elliptic curve, so $(5),(6)$ has infinitely many primitive solutions. 
Thus, one small solution to the first three conditions is then,
$$a,\;b,\;c,\;d = 216,\; 337,\; 140,\; 375$$
$$f,\;h =183, \;393 $$ 
The problem is the fourth condition which becomes,
$$a^2+b^2=(p^2+q^2)(r^2+s^2) = w^4\tag7$$
A third quadratic that is to be made not just a square but a fourth power, makes it highly unlikely that $(5),(6),(7)$ can be solved simultaneously. 
A: Select a pythogorean triplet of your choice.square root the answer.multyply each x and y with the answer.say x is 3 , y is 4 .the square root is 5.then (3×5)squared plus(4×5)squared equals 5^4.two squares equal to a fourth power
