a) Prove that if and 4|n, then x, y, and z are even.
b) Prove that if n is of the form , then n is NOT the sum of three squares.
How can we prove part a?
For part b, this is what I've got so far.
Suppose n is a sum of three squares (aim for a contradiction). Assuming the result of part a, since 4| , x,y, and z must be even, so x/2, y/2, z/2 are integers.
=> , so n/4 is also a sum of three squares
How to finish the proof from here?
Any help is appreciated!
[note: also under discussion in math links forum]
Suppose x,y,z are not all even.
To continue, 4|(y^2 + z^2) => y^2 + z^2 is even, so it must be that both of y,z are odd OR both are even. But we assumed that not all x,y,z are even, so it must be thiat y,z are odd.
So x is even, y odd, z odd, is this the ONLY case possible?
How to continue from here?
2) At the end, we get that 8k+7=7(mod8), and on the other hand, since 8k+7 is a sum of three squares, we get that 8k+7=/=7(mod8). This is where the contradiction comes, am I right?
4|(y^2 + z^2). Take cases now
Both y,z are even - we are done. Nothing more to be proved.
One of y,z is odd - sum is odd => above won't be true.
Both y,z are odd - sum is even but convince yourself that it won't be divisible by 4. Hence this possibility is also ruled out.
The case x even, y odd, z odd means that
x=0(mod 2), y=1(mod 2), z=1(mod 2)
But to look at divisibility by 4, we need to use mod 4. How to go from mod 2 to mod 4? (In general, I'm pretty confused about problems relating to a change of modulus)
I hope someone can explain this process of changing modulus.
Then is even which implies either are all even or one is even and two odd.
If all are even then all is done, so suppose one even (and without loss of generality let it be ) and the others odd. As and we must have .
Now as and are odd we may write:
for some . This is not divisible by , a contradiction ...