What?! $\left\lfloor a \right\rfloor\cdot\left\lfloor b \right\rfloor$ is obviously smaller than $\left\lfloor ab \right\rfloor.$ I am surprised this is a TST problem...

Master_of_Aops wrote: What?! $\left\lfloor a \right\rfloor\cdot\left\lfloor b \right\rfloor$ is obviously smaller than $\left\lfloor ab \right\rfloor.$ I am surprised this is a TST problem... Actually, they can be equal if the fractional part is very small. For example, wolfram’s result.

let $x = m\sqrt{2}$ $y = n\sqrt{7}$ $x= a+r$ $y = b+s$ where $a,b$ positive int $0\leq r< 1$ , $0 \leq s <1$ $\lfloor x \rfloor = a$ $\lfloor y \rfloor = b $ $\lfloor x y \rfloor = \lfloor ab+as + br +sr \rfloor = ab +\lfloor as + br +sr \rfloor.$ $ab \leq ab +\lfloor as + br +sr \rfloor.$ $0 \leq \lfloor as + br +sr \rfloor.$

I think StarLex1's solution would have been perfect if there was explanation for why $as+br+sr$ cannot be 0 (although kind of trivial). Anyways, nice solution . I also wonder how did this come out in TST .

Here's how to prove that $\lfloor as + br +sr \rfloor \neq 0$: $a^2+2ar+r^2=2m^2\implies r(2a+r)\in\mathbb{Z}^+\implies (2a+r)(s)\geq \frac{s}{r}\quad(*)$ $b^2+2bs+s^2=7n^2\implies s(2b+s)\in\mathbb{Z}^+\implies (2b+s)(r)\geq \frac{r}{s}\quad(**)$ $(*)+(**)\implies 2(as+br+sr)\geq \frac{s}{r}+\frac{r}{s}\geq 2\implies as+br+sr\geq 1\implies \lfloor as+br+sr\rfloor >0$ Solved by G. who wanted to remain anonymous

Quidditch wrote: Here's how to prove that $\lfloor as + br +sr \rfloor \neq 0$: $a^2+2ar+r^2=2m^2\implies r(2a+r)\in\mathbb{Z}^+\implies (2a+r)(s)\geq \frac{s}{r}\quad(*)$ $b^2+2bs+s^2=7n^2\implies s(2b+s)\in\mathbb{Z}^+\implies (2b+s)(r)\geq \frac{r}{s}\quad(**)$ $(*)+(**)\implies 2(as+br+sr)\geq \frac{s}{r}+\frac{r}{s}\geq 2\implies as+br+sr\geq 1\implies \lfloor as+br+sr\rfloor >0$ Solved by G. who wanted to remain anonymous very nice job!