prove or disprove that every positive integer is the sum of a perfect square and a prime number or 1 I am absolutely clueless on how to prove this statement and what makes it more difficult is not knowing if it is true or false to try and find a way to prove it. Tried numbers up to 20 and it was true for all but that is not a proof that it is true for all integers. How do i go about answering this question?
 A: Primes are positive.
So, 25 is not so expressible.  The sequence of non-expressible integers begins 25,34,58,64,85,91,121,130,169,196,214,289,324,... and is A014090 at the OEIS.
There, Dean Hickerson gives a proof that this sequence is infinite.  Note that the allowance of the number 1 as an honorary prime does not cause difficulty, since a square cannot differ by 1 from a square.
Added: Dean's proof goes like this.  Let $n$ be a positive integer such that $2n-1$ is composite.  Then if we can write $n^2$ as the sum of a square and a prime, we'll have
$$ n^2 = m^2 + p$$ for some integer $m$ and some prime $p>0$.  Then, 
$$ p = n^2-m^2 = (n-m)(n+m).$$  Since $p$ is prime, we must have 
$$ n-m=1$$
and so $p=n+m=2n-1$, which contradicts the primality of $p$.  So $n^2$ is not so representable.
Now, if we allow negative primes, we simply modify this and consider positive $n$ such that $2n-1$ and $2n+1$ are composite.  Then if $n^2=m^2+p$ for some integer $m$ and some prime $p$, possibly negative, we have
$$ p =n^2-m^2 = (n-m)(n+m)$$ and so $n-m=\pm 1$ and $n+m=2n+1$ or $n+m=2n-1$.  Since both values are composite, this contradicts $p$'s primality.  So $n^2$ is not so representable.
Now, there are infinitely many such $n$. Consider, for example, $n$ of the form 
$$ n = 13+15r$$ for $r$ a positive integer.  Then
$$ 2n-1 = 25+30r = 5(5+6r)$$
and
$$2n+1 = 27+30r = 3(9+10r)$$
are both composite, and so there are infinitely many integers not representable as the sum of a square and a prime (positive or negative).
