CatalinBordea wrote: Its floor surpasses $ 3/4. $ What does this mean?

DynamoBlaze wrote: CatalinBordea wrote: Its floor surpasses $ 3/4. $ What does this mean? Its integer part is greater than $ 3/4. $

Edit: Didn't notice q≤p.

Hunter01234 wrote: CatalinBordea wrote: Let be two natural primes $ 1\le q \le p. $ Prove that $ \left( \sqrt{p^2+q} +p\right)^2 $ is irrational and its floor surpasses $ 3/4. $ I think the problem is wrong. Check for p=2 and q=5. Your example does not satisfy $ q\le p. $

It's a typo in the statement. It should be fractional part instead of floor.

The first part of the question was covered by NikolsLife. Now $\left(\sqrt{p^2+q}+p\right)^2=4p^2+2q-r,$ where $r=\left(\sqrt{p^2+q}-p\right)^2.$ But $0<r<\frac{1}{4},$ where the LHS inequality is obvious, and the RHS one is equivalent to $q<p+\frac{1}{4}.$ Therefore, $\frac{3}{4}<\left\{\left(\sqrt{p^2+q}+p\right)^2\right\}<1.$

For the second part, we just need to show that the fractional part of $2p\sqrt{p^2+q}$ exceeds $\frac{3}{4}$. But $2p^2+q-1<2p\sqrt{p^2+q}<2p^2+q$, so it remains to be shown that $2p^2+q-\frac{1}{4}<2p\sqrt{p^2+q}$, which is trivial after squaring.

Sorry to break it to you fellas, but I'm afraid p and q aren't primes However, there isn't much of a deal of this problem anyway. The proof with the fractional part remains the same, or I couldn't find any problem with it so far. The slight change is in the irrational proof. Using the fact that p and q are natural and that p>=q, we easily see that p^2+q > p^2 and p^2+q < p^2+2*p+1 = (p+1)^2. So yeah, no biggie with proving that p^2+q isn't a square, so the whole thing is irrational.