Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?

I'll borrow newzad's caclulations and we'll prove that that last equation can't be a square number. We have:
$$a^2 - b^2 = n^2(k^4 - 6k^2 + 1)$$
Because $n^2$ is a square of some natural number $n$ we need to prove that $k^4 - 6k^2 + 1$ is a square number in order RHS to be also a square. First make a substitution $k^2 = t$ to reduce it to quadratic equation. And because we want it to be a square, set it to $m^2$. So we have:
$$k^4 - 6k^2 + 1 = t^2 - 6t + 1 = m^2$$
$$t^2 - 6t + (1 - m^2) = 0$$
Now solve this quadratic equation with respect to $t$. Because all coefficients are integers it'll have a integer solution, if the discriminant is a square number so we have:
$$D = b^2 - 4ac = (-6)^2 - 4(1-m^2) = 36 - 4 + 4m^2 = 2^2(9 - 1 + m^2) = 2^2(m^2 + 8)$$
Again we have that the discriminant will be a square number if $m^2 + 8$ is a square so we have:
$$m^2 + 8 = l^2$$
$$m^2 - l^2 = -8$$
$$(m-l)(m+l) = -8$$
By checking for factors of $-8$ we get that $m=\pm1$ and $l=\pm3$ is the only integer solutions. 
So we can subtitute back in the initial equation and we have:
$$t^2 - 6t + 1-1 = 0$$
$$t^2 - 6t = 0$$
$$t(t-6) = 0$$
$$t_1 = 0 \quad \quad \quad t_2 = 6$$
This implies that:
$$k_{1/2} = 0 \quad \quad \quad k_{3/4} = \pm\sqrt{6}$$
But none of this is possible, because in newzad's caculation we have:
$$\frac{a}{b} + \frac{c}{b} = k$$
Because $a,b,c \in \mathbb{N}$ it's impossible for $k=0$ and also because $a,b,c$ are positive integers we have that $\frac{a}{b}$ and $\frac{c}{b}$ are rational numbers, but we know that a sum of two rational numbers, will never be an irrational numbers, so it impossible $k = \pm \sqrt{6}$
Because there aren't any more possibilities for $k$ we have that it's impossible for $a,b,c,d \in \mathbb{N}$ the following to hold:
$$a^2 + b^2 = c^2 \quad \quad \quad a^2 - b^2 = d^2$$
Q.E.D.
