Question

Putnam help here. Question attached on next post.

Answer 1

In order to show that there are infinite n, I need to show that there are infinitely many integers p and q such that:\[p^{2} + q^{2} = 2n^{2} + 1\]I might be somewhat sleepy, but I didn't have any idea to prove it. Should I try to contradict it?

Answer 2

Also, if anyone has ideas for solving the problem, feel free to post it, I'd like different ideas to prove it. :-)

Answer 3

However unlikely it seems that this would be the correct approach, we could try a proof by contradiction by listing all n that satisfy this, and then creating some number out of those n that also satisfies this property.

Answer 4

Got it, I think.
It's plausible to state that, suppose that there aren't infinitely many p and q. Let P and Q be the largest values possible. So, we have:\[P^{2} + Q^{2} = 2N^2 + 1\]But N is not the largest possible for n, as we can write, at least: M = N^2 + N^2. Add one, and we will get:
M + 1 = 2N^2 + 1, that would satisfy a larger p and q, which is a contradiction.

Answer 5

Does this hold?

Answer 6

Maybe "larger" is the incorrect word. Maybe, "last" possible would be more accurate.

Answer 7

Honestly, I'm not sure if I follow why that would work. It seems to me as if we're getting the same p and q.

Answer 8

Yeah, that's was a tautology. I will try to rework it.

Answer 9

If n = 1
This wouldn't work :/
n+2 = 3
if p is an integer, q wouldn't be one

Answer 10

I think it suffices if p and q are any numbers. I only need to prove that they are a infinite set.

Answer 11

Could we use the identity \((a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2\)? Using this, we can create a new pair of p and q.

Answer 12

I will try to give more insight into my proof. Suppose A is the set of all n that satisfies the condition. We have that:\[2n^{2} + 2 = (n+1)^{2} + (n-1)^{2}\] for all n in A. From the LHS, we have:\[(2n^2 + 1) + 1\]If (2n^2 + 1) has the form p^2 + q^2, I think I will be near the proof.

Answer 13

Actually, no. I got it.

Answer 14

From the Pell equation,\[x^2 - 2y^2 = 1\]Let n = 2y^2. We have that n = y^2 + y^2, n + 1 = x^2 + 0^2, n + 2 = x^2 + 1^2. The equation has infinite solutions, therefore, there must be infinite n. Right?

Answer 15

That looks like it should work.

Answer 16

Thanks for the help @KingGeorge and others. I will try to solve it by contradiction also, it seems it's feasible, only harder than a geometric argument.