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Does the equation $$x^2+y^2=1$$ have a solution $(x,y)$ in the natural numbers? If we assume natural numbers do not include $0$, can this equality hold? I assumed it is false, since only $1$, $-1$ and $0$ give that solution but $0$ and $-1$ cannot be used. So there is no way it can be true. Is this correct?

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  • $\begingroup$ Try to look for solutions such that $x\ge 19$. $\endgroup$ – Saucy O'Path Jun 12 '18 at 7:35
  • $\begingroup$ Clearly false. Putting minimum value of this expression is 2. $\endgroup$ – Lord KK Jun 12 '18 at 7:35
  • $\begingroup$ what do you mean look for solutions above 19 $\endgroup$ – Patrick Jun 12 '18 at 7:40
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No. Simply put, $$x^2=1-y^2$$ Therefore, $x=\pm \sqrt{(1-y^2)}$. Talking about natural numbers lets keep $x=+\sqrt{(1-y^2)}$ from the first quadrant and therefore add another condition $x\geq 0$ and $y \geq 0$.

Since root implies that for real solutions \begin{align} & 1-y^2 \geq 0 \\ \implies & -1 \leq y \leq 1 \end{align}

The intersection of this condition with $y\geq 0$ gives $y \leq 0 \leq 1$.

Similarly, $0 \leq x \leq 1$. Therefore if $x$ and $y$ has to take natural values then $x$ and $y$ have to be 1 but then they wouldn't lie on our circle.

Hence why the statement is false.

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  • $\begingroup$ Welcome to MSE. It is in your best interest that you use MathJax. $\endgroup$ – José Carlos Santos Jun 12 '18 at 7:57
  • $\begingroup$ Thanks for the advice and warm welcome. $\endgroup$ – Pi_die_die Jun 12 '18 at 8:16
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Just to give some geometric interpretation to the answer of Vidhu.

The equation $x^2+y^2=1$ describes the set of point in the unit circle in the plane. By drawing the circle you can easily find all natural (or even integer) solutions.

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