fermats inifinite descent problems

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September 3rd 2008, 10:57 AM
Proggy

fermats inifinite descent problems

I am not even real sure what the second problem is actually asking me to do. Any help would be appreciated. All eight problems of this last chapter in my Intro to Number Theory class seem incredibly hard to me. I will post a reply with my hopeful solution to the first problem which I think I have at least partially figured out.

#1 (solved)
Show that the equation

$x^4 + 4y^4 = z^2,\ \ \ \ \ x \neq 0, y \neq 0, z \neq 0$

has no solutions. It may be helpful to reduce this to the case that x > 0, y > 0, z > 0, (x,y)=1, and then by dividing by 4 (if necessary) to further reduce this to where x is odd.

#2
Show that there is no right triangle with integral sides whose area is a perfect square by showing that it suffices to work with primitive triangles and with them, one is led to the equation given in problem 1.

and #3 (solved)...I printed out the latex help sheet, but i do not see a 'not equal' sign anywhere on it. Or a not congruent sign for that matter. I typed it out for now, but can anyone tell me what is proper to use for that? I will search the latex forum for it later.
September 3rd 2008, 11:19 AM
Proggy

$x^4 + 4y^4 = z^2$
$(x^2)^2 + (2y^2)^2 = z^2$

this is a pythagorean triplet, so
$2y^2 = 2pq,\ x^2 = p^2 - q^2,\ z = p^2 + q^2$
$2y^2 = 2pq$
$y^2 = pq = u^2v^2$
$p = u^2,\ q = v^2$

$x^2 = p^2 - q^2$
$x^2 + q^2 = p^2$
this is a pythagorean triplet, so
$x = r^2 - s^2, q = 2rs, p = r^2 + s^2$

$q = v^2 = 2rs = g^2h^2$
$2r = g^2$
$r = (g^2)/2$
$s = h^2$

$p = r^2 + s^2$
substituting gives
$u^2 = [(g^2)/2]^2 + (h^2)^2$
$u^2 = (g^4)/4 + h^4$
for $u < u^2 = p < p^2 + q^2 = z$ or $u < z$

so somewhere I am messing up and end up dividing by 4 instead of multiplying by 4. Not sure what I am doing wrong tho. I am trying to investigate the hint to divide by 4 (if necessary) and that I am not looking at even vs odd. I do know that I have two cases, where x and z are even and y is odd, and when x and z are both odd and y is either even or odd.
September 3rd 2008, 12:07 PM
Laurent

You proved that $g^2=2r$ , hence $g$ is even. It writes as $g=2g'$ so that $g^4=16g'^4$ , and the final equation becomes: $u^2=4g'^4+h^4$ . This is what you need, isn't it ?

Laurent.
September 3rd 2008, 12:17 PM
Proggy

Quote:

Originally Posted by Laurent

You proved that $g^2=2r$ , hence $g$ is even. It writes as $g=2g'$ so that $g^4=16g'^4$ , and the final equation becomes: $u^2=4g'^4+h^4$ . This is what you need, isn't it ?

Laurent.

yes Laurent, that is perfect for finishing up #1, which becomes
$x^4 + 4y^4 = z^2$
$(x^2)^2 + (2y^2)^2 = z^2$

this is a pythagorean triplet, so
$2y^2 = 2pq,\ x^2 = p^2 - q^2,\ z = p^2 + q^2$
$2y^2 = 2pq$
$y^2 = pq = u^2v^2$
$p = u^2,\ q = v^2$

$x^2 = p^2 - q^2$
$x^2 + q^2 = p^2$
this is a pythagorean triplet, so
$x = r^2 - s^2, q = 2rs, p = r^2 + s^2$

$q = v^2 = 2rs = g^2h^2$
$2r = g^2$ , so g is even which means $2r = g^2 = (2a)^2 = 4a^2$
$r = 2a^2$
$s = h^2$

$p = r^2 + s^2$
substituting gives
$u^2 = (2a^2)^2 + (h^2)^2$
$u^2 = 4a^4 + h^4$
for $u < u^2 = p < p^2 + q^2 = z$ or $u < z$

and I also did some deductive reasoning on the latex hint sheet and saw the \geq, so tried \neq and it worked. So now I just need someone to translate #2 into a form that I understand so that I know what to do with it.
September 3rd 2008, 12:34 PM
Laurent

About the triangle, here's (more than) an hint: let $a$ and $b$ be the perpendicular sidelengths, and $c$ be the length of the other side. Then $a^2+b^2=c^2$ and the area $\frac{ab}{2}$ is a square, hence $ab=2d^2$ for some integer $d$ .
If you have shown that you may suppose that $a$ and $b$ are relatively prime and that $a=2a'$ is even, then you can deduce from $ab=2d^2$ that $a'$ and $b$ are squares. So that the equation $a^2+b^2=c^2$ may be written as $x^4+4y^4=z^2$ for some $x,y,z\in\mathbb{N^*}$ .

For the LaTeX question, you can use "\neq" to get the $\neq$ symbol.

Laurent.