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?
 A: $${ \left( \frac { a }{ b }  \right)  }^{ 2 }+1={ \left( \frac { c }{ b }  \right)  }^{ 2 }\\ \frac { c }{ b } +\frac { a }{ b } =k\\ \frac { c }{ b } -\frac { a }{ b } =\frac { 1 }{ k } \\ \frac { c }{ b } =\frac { { k }^{ 2 }+1 }{ 2k } \\ \frac { a }{ b } =\frac { { k }^{ 2 }-1 }{ 2k } \\ $$Say:$$a=({ k }^{ 2 }-1)n\\ b=2kn\\ { a }^{ 2 }-{ b }^{ 2 }={ n }^{ 2 }({ k }^{ 4 }-6{ k }^{ 2 }+1)=n^2(k^2-3)^2-8=d^2$$
Is the result a perfect square?
A: Dividing by $(a,b)$, we can assume that $(a,b)=1$.
Multiplying the two equations yields
$$
a^4-b^4=c^2d^2\tag{1}
$$
whereby
$$
b^4+(cd)^2=a^4\tag{2}
$$

This answer characterizes all primitive Pythagorean Triples; that is, there must be a pair of positive integers $(m,n)$ so that $(m,n)=1$ and $m+n$ is odd so that
$$
x=m^2-n^2,\quad y=2mn,\quad z=m^2+n^2\tag{3}
$$
Suppose that $(x,y,z)$ is the Pythagorean triple with the smallest hypotenuse so that at least two sides are a square or twice a square.
Suppose that $z$ and $x$ are squares (since they are odd, they can't be twice a square). Then $(\sqrt{xz},n^2,m^2)$ has a smaller hypotenuse.
Suppose that $z$ is a square and $y$ is a square or twice a square. Then $m$ and $n$ must both be either a square or twice a square. Then $(m,n,\sqrt{z})$ has a smaller hypotenuse (since $x,y\ge1$, $z\gt1$).
Suppose that $x$ is a square and $y$ is a square or twice a square. Then $m$ and $n$ must both be either a square or twice a square. Then $(\sqrt{x},n,m)$ has a smaller hypotenuse.
Thus, there is no primitive Pythagorean Triple with at least two sides a square or twice a square.
If $(2)$ were true, there would be a primitive Pythagorean Triple with two sides a square.
A: 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.
A: Summing and subtracting, you get $$\begin{cases} 2a^2=c^2+d^2\\2b^2=c^2-d^2\end{cases}$$ Thus $c,d$ have the same parity. If they are both even, we obtain $2a^2=4c'^2+4d'^2$ and $2b^2=4c'^2-4d'^2$, so $$\begin{cases} a^2=2(c'^2+d'^2)\\b^2=2(c'^2-d'^2)\end{cases}$$
Since $2\mid a^2,b^2$, we must have $a^2=4a'^2$, $b^2=4b'^2$, so we get that  $$\begin{cases} 2a'^2=c'^2+d'^2\\2b'^2=c'^2-d'^2\end{cases}$$
And now you're back to the same, but with smaller numbers. And thus we descend. 
It remains the case that both are odd, which I couldn't solve.
