# Integer Number Theory - n = p + a^2

1. Feb 13, 2013

### jj7964130

1. The problem statement, all variables and given/known data
Prove or disprove: If n is a positive integer, then $n=p+a^2$ where
• $a\in\mathbb{Z}$
• $p$ is prime or $p=1$
Note that the interpretation of "prime" used here includes negative primes. So, an exhaustive list of possibilities for p is $p=1,\pm2,\pm3,\pm5,\pm7,\pm11,\cdots$

2. Relevant equations
Prime is defined such that if p prime and p divides the product ab, then either p divides a or p divides b. Also, primes are irreducible.

Additionally, the fundamental theorem of arithmetic that defines all integers as a unique product of positive primes may be useful.

3. The attempt at a solution
Originally, I had found 25 as a counterexample that cannot be written as the sum of a prime and a square. Then the problem was clarified to include negative primes, and I'm a bit lost as to where I should start. Namely, I'm not sure if I should be working towards proving or disproving the argument. If anyone's worked through this already and could send me some guidance in the right direction, it would be appreciated.

2. Feb 13, 2013

### Staff: Mentor

$25=-11+6^2$

I see a nice counterexample with a very small n.

3. Feb 13, 2013

### jj7964130

I'm not sure if an example is necessarily the best route to take....

Regardless of small n, it's very difficult to take into account all negative primes of large magnitude and squares of equally large magnitude. Since the list of possibilities is infinite, I don't see a concrete way to say "this will never be true."

4. Feb 13, 2013

### Staff: Mentor

Hmm, I did not see "positive" in the requirements. Ok, 0 does not work.
[STRIKE]There might be (p,a) for every positive n.[/STRIKE]

Last edited: Feb 13, 2013
5. Feb 13, 2013

### Dick

You do want to concentrate on finding a counterexample. Here's a hint: look at the case where n is a perfect square. Do you see how that helps in limiting the infinity of options?