MHB Sum of Two Squares: Can $5^{64}-3^{64}$ Be Written?

[anemone]

Is it possible to write $5^{64}-3^{64}$ as the sum of two squares?

 

[kaliprasad]

Is it possible to write $5^{64}-3^{64}$ as the sum of two squares?

Spoiler

$5^{64}-3^{64}$
=$(5^{32}+3^{32})(5^{16}+3^{16})(5^8+3^8)(5^4+3^4)(5^2+3^2)(5+3)(5-3)$
= $16((5^{32}+3^{32})(5^{16}+3^{16})(5^8+3^8)(5^4+3^4)(5^2+3^2)$
= $4^2((5^{32}+3^{32})(5^{16}+3^{16})(5^8+3^8)(5^4+3^4)(5^2+3^2)$

as product of 2 numbers both sum of 2 squares can be represented as sum of 2 squares and 4^2 is a square so argument repeatedly we have the ans Yes

 

[topsquark]

Spoiler

I'm not familiar with this theorem. Are you saying that any form [math](a^{2p} + b^{2p})(c^{2q} + d^{2q} )(e^{2r} + f^{2r})[/math] can always be written as the sum of two squares? Or do we additionally need c = e =a, d = f = b or something?

-Dan

 

[Opalg]

Spoiler

I'm not familiar with this theorem. Are you saying that any form [math](a^{2p} + b^{2p})(c^{2q} + d^{2q} )(e^{2r} + f^{2r})[/math] can always be written as the sum of two squares? Or do we additionally need c = e =a, d = f = b or something?

-Dan

[sp]
The identity $(a^2+b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2$ shows that a product of sums of two squares is also a sum of two squares.
[/sp]

 

[topsquark]

(Doh) I knew that one.

Thanks!

-Dan

 

[anemone]

Thanks Kali for your solution!

And thanks to Opalg too for explaining thing for topsquark! I appreciate that!