Proving 2n Representable When n is: Converse True?

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Homework Statement


Prove that 2n is representable when n is. Is the converse true?
Representable is when a positive integer can be written
as the sum of 2 integral squares.

The Attempt at a Solution


so n can be written as x^2+y^2
x and y are positive integers
so then 2n=2(x^2+y^2)
I am not really sure where to go next, maybe i should look a the prime factors.
Just to make sure that i know what converse is,
Would the converse be if 2n is representable so is n.
 
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cragar said:

Homework Statement


Prove that 2n is representable when n is. Is the converse true?
Representable is when a positive integer can be written
as the sum of 2 integral squares.

The Attempt at a Solution


so n can be written as x^2+y^2
x and y are positive integers
so then 2n=2(x^2+y^2)
I am not really sure where to go next, maybe i should look a the prime factors.
Just to make sure that i know what converse is,
Would the converse be if 2n is representable so is n.

Brahmagupta–Fibonacci identity. Look it up. And yes, that would be the statement of the converse.
 
cragar said:

Homework Statement


Prove that 2n is representable when n is. Is the converse true?
Try some specific examples, like 25 = 32 + 42

or 29 = 22 + 52 .

Just to make sure that i know what converse is,
Would the converse be if 2n is representable so is n.
Yes, that's the converse.
 
ok thanks for the responses. Knowing my teacher I would need to prove the
Brahmagupta–Fibonacci identity, but I guess I could multiply it out and show that
the left side equaled the right side. I think the converse is true but ill think about how to prove it.
 
cragar said:
ok thanks for the responses. Knowing my teacher I would need to prove the
Brahmagupta–Fibonacci identity, but I guess I could multiply it out and show that
the left side equaled the right side. I think the converse is true but ill think about how to prove it.

Yes, you should definitely think much harder about the converse. And sure, it's much easier to prove them than to discover they exist.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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