TITLE: $p+q\neq 1+pq$ for distinct odd primes $p$ and $q$
QUESTION [5 upvotes]: I'm trying to show that that $\sqrt p + \sqrt q$ cannot be written as a linear combination of $1$ and $\sqrt{pq}$ with rational coefficients, and I have boiled it down to showing that $p+q \neq 1+pq$ when $p$ and $q$ are distinct primes.
I tried doing this by contradiction and showed that if, say, $p=2$ then $2+q=1+2q \implies q=1$, which is a contradiction because $q$ is prime.
I would appreciate if you could show me how to deal with the case where both $p$ and $q$ are odd.
(I've mostly tried modding out $p$ or $q$ and moving things around, but I'm not really getting anywhere with that.)

REPLY [7 votes]: If $p = 2j + 1$ and $q = 2k + 1$, then $p + q = 2j + 2k + 2$ while $1 + pq = 1 + (2j + 1)(2k + 1) = 4jk + 2j + 2k + 2$.
Therefore $(pq + 1) - (p + q) = 4jk > 0$.