Does the equation $x^2 + y^2 = 1$ have a solution in natural numbers?
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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$\begingroup$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.
$\endgroup$ 2 $\begingroup$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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