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Solve the following system of simultaneous equations: $$ \left \{ \begin{matrix} x^2+[y]=10 \\ y^2+[x]=13 \end{matrix} \right . $$

where square brackets $[...]$ denote integral part of a number (a.k.a. floor)

This is what I've come to:

Case I. $y\ge 0$.

First equation yields: $|x|\le\sqrt{10}$ (|..| denotes absolute value), therefore $-4\le [x] \le 3$. Substitution into the second equation gives: $10\le y^2 \le 17$, so $3\le [y] \le 4$.

Then, from the first equation it can be evaluated that $x=\pm\sqrt{7}$ for $[y]=3$ and $x=\pm\sqrt{6}$ for $[y]=4$.

If $x>0$ (i.e $x=+\sqrt{6}$ or $x=+\sqrt{7}$), then $[x]=2$, so (from 2-nd equation) $y=\sqrt{11}$ and $[y]=3$. This gives a solution: ($\sqrt{7}$,$\sqrt{11}$).

If $x<0$ (i.e $x=-\sqrt{6}$ or $x=-\sqrt{7}$), then $[x]=-3$, so $y^2=16$ and $[y]=y=4$. This gives another solution: ($-\sqrt{6}$, $4$),

Case II. $y \lt 0$, $x \ge 0$.

From second equation $-\sqrt{13} \le y \lt 0$, therefore $-4 \le y \le -1$, so from the first equation $\sqrt{11} \le x \le \sqrt{14}$ and $[x]=3$. Then from second equation $y=-\sqrt{10}$, so $[y]=-4$ and $x=\sqrt{14}$. Hence the third solution: ($\sqrt{14}$,$-\sqrt{10}$).

Case III. $x \lt 0$, $y \lt 0$.

Any ideas?

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Hint: $\;\lfloor x \rfloor \le x \lt \lfloor x \rfloor + 1\,$ and the same for $\,y\,$, then:

$$ \left \{ \begin{matrix} 10 \le x^2+y \lt 11 \\ 13 \le y^2+x \lt 14 \end{matrix} \right . \;\;\implies\;\; 23 \;\le\; x^2+y+y^2+x \,=\, (x+1/2)^2+(y+1/2)^2 - 1/2 \;\lt\; 25 $$

It follows that $|x+1/2|, |y+1/2| \lt \sqrt{25+1/2}\,$ which gives useful bounds.

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  • $\begingroup$ The solution I was missing is ($-\sqrt{15}$,$-\sqrt{17}$). Now, it's all clear! $\endgroup$ – cyanide Jan 14 '18 at 8:25
  • $\begingroup$ @cyanide That's the one, indeed. Well done. $\endgroup$ – dxiv Jan 14 '18 at 23:53

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