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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.)

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    $\begingroup$ $p+q=1+pq \Rightarrow p-pq=1-q\Rightarrow p(1-q)=1-q$ $\endgroup$ – lulu Aug 10 '15 at 3:14
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    $\begingroup$ So for $p,q$ primes, you need to show that $(p-1)(q-1)\neq 0$, is that right? $\endgroup$ – abiessu Aug 10 '15 at 3:14
  • $\begingroup$ Thanks to both @lulu @abiessu! I can't believe this was that easy :P $\endgroup$ – a girl Aug 10 '15 at 3:18
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    $\begingroup$ It is surprising how often $(x-a)(y-b) = xy - ay -bx + ab$ comes in handy. $\endgroup$ – marty cohen Aug 10 '15 at 5:15
  • $\begingroup$ @abiessu Connect a few dots and you've got the answer sophia ought to accept. $\endgroup$ – James47 Aug 10 '15 at 20:51
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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 $(pq + 1) - (p + q) = 4jk > 0$.

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