Need help with the proof of the following theorem of the equation $x^4+y^4=z^4$ Theorem 7-2. The equation $x^4+y^4=z^4$ is not solvable in nonzero integers.  
Proof: It suffices to show that there is no primitive solution of the equation $$x^4+y^4=z^2$$ (why?)
Suppose that $x,y,z$ constitute such solution; with no loss in generality we may take $x>0$, $y>0$, $z>0$, and y even. Writing the supposed relation in the form
$$(x^2)^2+(y^2)^2=z^2$$
we see from Theorem 7-1 (the theorem of Pythagorean triples ) that $x^2=a^2-b^2$, $y^2=2ab$,$z=a^2+b^2$ where $(a,b)=1$ and exactly one of a an b is odd. If a were even, we would have $$1 \equiv x^2=a^2-b^2 \equiv -1(mod \text{ } 4)$$
(how is the latter true)so b is even. We apply theorem of Pythagorean triples again , this time to the equation $$x^2+b^2=a^2$$
(why do we do the latter?) and obtain $x=p^2-q^2$,$b=2pq$, $a=p^2+q^2$, where $(p,q)=1$, $p>q>0$, and not both p an q are odd. From $y^2=2ab$ we have $$y^2=4pq(p^2+q^2)$$ 
Here any two of p, q and $p^2+q^2$ are relatively prime and hence each must be a square (why are they relatively prime and why must each of them be a square?): $p=r^2$, $q=s^2$, $p^2+q^2=t^2$, whence $t>1$ and $$r^4+s^4=t^2$$
Now $x=r^4-s^4$, $y=2rst$, $z=a^2+b^2=r^8+6r^4s^4+s^8$, so that $$z>(r^4+s^4)^2=t^4$$
or $t<z^{1/4}$. It follows that if one nonzero solution of $x^4+y^4=z^2$ were known, another solution $r,s,t$ could be found for which $rst \neq 0$ and $1<t<z^{1/4}$. If we started from $r,s,t$ instead of $x,y,z$, a third solution $r',s',t'$ could then be found such that $1<t'<t^{1/4}$, and so on. But this would yield an infinite decreasing sequence of positive integers, $z,t,t',....,$ which is impossible. Thus there is no nonzero solution. (Can you explain this better?)
Thanks for your help
 A: If $(x,y,z)$ is a solution to $x_1^4+x_2^4=x_3^4$, then $(x,y,z^2)$ is a solution to $x_1^4+x_2^4=x_3^2$.  This is the same thing as saying if $x_1^4+x_2^4=x_3^2$ has no solution, then neither does $x_1^4+x_2^4=x_3^4$.  
$1\equiv x^2 \equiv a^2-b^2 \equiv 0-1 \equiv -1 \mod 4$, because $a$ is even and $b$ is odd, which cannot be the case.  
We know $x^2=a^2-b^2$, so naturally $x^2+b^2=a^2$, for which we can apply the same trick again.  
$p$ and $q$ are coprime by construction.  Whenever you have a primitive Pythagorean triple, the terms being added are coprime.  $p=r^2, q=s^2$ and $p^2+q^2=t^2$ because they are all coprime factors of a perfect square $y^2$.  
The last paragraph is just saying that we could consistently construct $z>t>t'>t''>t'''>\ldots$, all of which need to be positive whole numbers, which is not possible.  That should be pretty intuitive.  For example, consider the sequence $64, 32, 16, 8, \ldots$.  If they are all positive whole numbers, then it should be pretty clear that this sequence needs to end.  Positive ensures you can't go into negatives, and whole numbers ensures that the sequence can't start squishing together.  
