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Prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z)=1$.

Do I use the terms $x= r^2 - s^2$, $y = 2rs$, and $z = r^2 + s^2$ to prove this problem?

Thanks for any help.

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While it is true that any solution can be expressed with $x=r^2-s^2$, $y=2rs$, and $z^2=r^2+s^2$ (note the square in $z$) because $x$, $y$, and $z^2$ are a primitive pythagorean triple, not every pair of values of $r$ and $s$ (with $r\gt s$, $\gcd(r,s)=0$, and $r$ and $s$ of opposite parity) will yield a solution, since you need $r^2+s^2$ to be a square. But you can then implement the formula again to find solutions to $z^2=r^2+s^2$, and then... –  Arturo Magidin Jul 18 '12 at 19:14
    
Let $x+yi = (a+bi)^4$ –  Thomas Andrews Jul 18 '12 at 19:24
    
What do you mean with "$(x,y,z)=1$"? –  celtschk Jul 18 '12 at 20:23
    
@celtschk: It means $x,y,$ and $z$ are relatively prime. –  Brandon Carter Jul 18 '12 at 20:27

2 Answers 2

Hint Let $w=z^2$. Then

$$x^2+y^2=w^2$$

Solve this Pytagorean equation, and prove that infinitely many $w$ are perfect squares.

With your notation $w=r^2+s^2$ so you need $r,s$ to satisfy $r^2+s^2=z^2$...

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N.S I want to adapt your answer to prove $x^2 + y^2 = z^4$ has infinitely many "primitive" solutions with x,y,z natural numbers. I'm just unsure what you mean by "so you need r,s to satisfy $r^2 +s^2=z^2...$ " –  jakey Nov 16 '13 at 21:26
    
@user109323 The general primitive solution to the Pytagorean equation $x^2+y^2=w^2$ is $x=2rs, y=r^2-s^2$ and $w=r^2+s^2$. Now if you want $w=z^2$ you get $x=2rs, y=r^2-s^2$ and $z^2=r^2+s^2$. Since you need $z$ to be an integer, the equation $z^2=r^2+s^2$ must have integral solution. This is again the Pytagorean equation. Solve it, and then replace the $s$ and $r$ you get in $x=2rs, y=r^2-s^2$.... –  N. S. Nov 16 '13 at 23:06
    
I can't add comments so have to put this in an answer each time, rather than awkwardly try and discuss it here, I would be extremely thankful if you could check out my response here : math.stackexchange.com/questions/570373/… thanks! -user109323 with a more human username, Jakey –  jakey Nov 17 '13 at 13:16

Actually, a very strong statement is true, and quite easy: given any positive integer $n,$ such that all prime factors $p$ of $n$ satisfy $p \equiv 1 \pmod 4,$ then there is a solution to $$ x^2 + y^2 = n $$ with $$ \gcd(x,y) = 1.$$

For $n$ a prime (so itself $1 \pmod 4$) the existence of a solution was finally proved by Euler. After that, there is a simple mathematical induction, using the two choices in someone's formula (Brahmagupta with $N=-1$), anyway $$ (a^2 + b^2)(c^2 + d^2) = (ac+bd)^2 + (ad-bc)^2, $$ or $$ (a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2 $$ with the induction hypothesis $$ \gcd(a,b) = \gcd(c,d) = 1. $$

Indeed, given such an $n,$ with $r$ distinct prime factors and each $p_i \equiv 1 \pmod 4,$ but we otherwise ignore the exponents, the number of representations $$ x^2 + y^2 = n $$ with $$ 0 < x < y, \; \; \gcd(x,y) = 1 $$ is exactly $$ 2^{r-1}. $$ For example, about ignoring exponents, we get $$ 1 +4 = 5, $$ $$ 9+16 = 25 = 5^2, $$ $$ 4 + 121 = 125 = 5^3, $$ $$ 49 + 576 = 625 = 5^4. $$

Alright, nobody is paying attention to this one, but I thought I ought to make sure I knew how the induction proof went. It is easier if we demand the second factor prime, so we are taking $$ a^2 + b^2 = n, \; \; \gcd(a,b) = 1,$$ $$ c^2 + d^2 = p, \; \; p \equiv 1 \pmod 4, \; \; \mbox{of course} \; \; \gcd(c,d) = 1. $$ We are going to represent $np.$ One way is $$ np = (ac+bd)^2 + (ad-bc)^2. $$ Suppose these are not relatively prime, there is some positive prime $q$ such that $$ q | ac+bd \; \; \mbox{and} \; \; q | ad-bc. $$ Multiply the first by $d$ and the second by $c$ and subtract, we get $$ q | b (c^2 + d^2), \; \; \mbox{or} \; \; q | b p. $$ Back to our pair, multiply the first by $c$ and the second by $d$ and add, we get $$ q | a (c^2 + d^2), \; \; \mbox{or} \; \; q | a p. $$ Since $ \gcd(a,b) = 1, $ it follows that $q |p$ and then that $q=p.$ So the only possible GCD is a power of $p$ itself. If, in addition, $$ p | ac-bd \; \; \mbox{and} \; \; p | ad+bc, $$ we get $ p | 2ac $ and $p | 2bc,$ or $p | a$ and $p|b.$ This is false, as $\gcd(a,b) = 1.$

To put it briefly, the only possible obstacle to a proper representation of $np$ is a common factor of $p$ itself, but if $p$ divides all four of $ac+bd, \; ad-bc, \; ac-bd, \; ad+bc, $ then it divides both $a,b,$ a contradiction. So, at least one of the four given expressions is not divisible by $p,$ and the pair using that expression gives a proper/primitive representation of $np$ as the sum of two squares. Naturally, the other expression in the successful pair is also prime to $p.$

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