# Math Help - Diophantine equation

1. ## Diophantine equation

Hello everyone!
I have some trouble with this proof: prove that there are infinitely many solutions to the equation: $x^2 + y^2 = z^4 \mbox{, with g.c.d} (x,y,z) = 1$

2. Originally Posted by rubik mania
Hello everyone!
I have some trouble with this proof: prove that there are infinitely many solutions to the equation: $x^2 + y^2 = z^4 \mbox{, with g.c.d} (x,y,z) = 1$
let $a,b$ be any (positive) integers with $\gcd(a,b)=1$ and $2 \nmid a+b.$ put $x+yi=(a+bi)^4.$ then $x-yi=(a-bi)^4$ and hence $x^2+y^2=(a^2+b^2)^4.$

equating the real and complex parts in $x+yi=(a+bi)^4$ will gives us these solutions: $x=a^4 - 6a^2b^2 + b^4, \ \ y=4a^3b - 4ab^3,$ and $z=a^2+b^2.$

it's immediate that $\gcd(x,y,z)=1.$

3. Thanks!

For similar questions that involve proving that a Diophantine equation has infinite/finite/no solutions, what would be the general strategy?

4. Originally Posted by rubik mania
Thanks!

For similar questions that involve proving that a Diophantine equation has infinite/finite/no solutions, what would be the general strategy?
it's number theory ... there is almost never a general strategy! haha ... anyway, i just showed you one approach. two more:

1) reducing the equation modulo some integer or the method of "infinite descent". these methods are normally used to prove that an equation has no solutions.

2) investigating more sophisticated equations usually requires some knowledge of algebraic number theory or elliptic curves.