## FoolForMath Group Title Another fiasco of wolframalpha: How many integer solutions are there for the in-equation $x^2 +y^2 \le 25$? And this is what wolframalpha says: http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2+%3C%3D+25 But I got a different answer, Just for fun, can you find the right answer? 2 years ago 2 years ago

Obviously the maximum value $$x$$ or $$y$$ can take is $$5$$, and the minimum is $$-5$$. We can list all the choices as: At $$x=\pm 5$$, $$y$$ can take only the value $$0$$. At $$x=\pm4$$, $$y$$ can take the values $$0, \pm1, \pm2$$ and $$\pm3$$. At $$x=\pm 3$$, $$y$$ can take the values $$0, \pm1, \pm2, \pm3$$ and $$\pm4$$. At $$x=\pm 2$$, $$y$$ can take the value $$0, \pm 1, \pm 2, \pm 3$$ and $$\pm 4$$. At $$x=\pm1$$, $$y$$ can take the values $$0, \pm1, \pm 2, \pm 3$$ and $$\pm 4$$. At $$x=0$$ $$y$$ can take the values $$0, \pm1, \pm2, \pm 3, \pm 4$$ and $$\pm 5$$. So the total number of integer solutions is $$2+2(7)+2(9)+2(9)+2(9)+11=81.$$