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Problem: Let $$k$$ be a fixed odd positive integer. Find the minimum value of $$x^2 + y^2$$, where $$x,y$$ are non-negative integers and $$x+y=k$$.

Solution: We have $$y=k-x$$. Therefore we get an equation in $$x$$ where $$k$$ is a constant, precisely $$f(x) = x^2 + (k-x)^2$$.

To minimise, we differentiate $$f(x)$$ w.r.t $$x$$.

So $$f'(x) = 4x-2k = 0$$ (for minimum $$f(x)$$)

That gives us $$x=\frac{k}{2}$$.

But the question tell us that $$k$$ is odd and $$x$$ is an integer. therefore we have to take the closest possible integer value to $$\frac{k}{2}$$, which is $$\frac{k+1}{2}$$ and $$\frac{k-1}{2}$$.

As already defined, taking $$x$$ to be any one of the above $$y$$ automatically takes the other value.

Therefore the minimum value of $$x^2 + y^2$$ is given by $$(\frac{k+1}{2})^2 + (\frac{k-1}{2})^2$$ $$= \frac{k^2+1}{2}$$.