Show that $n + (\sqrt n)$ is not a perfect square

Question

Show that $n + (\sqrt n)$ is not a perfect square, where $n$ is a natural number and $(\sqrt n)$ is the nearest integer of $\sqrt n$.

Answer 1

Let $n=a^2+b$ with $0\leq b\leq2a$.

Then $(\sqrt n)=a$ or $a+1$.

• If $b\leq a$ then $(\sqrt n)=a$ because $$a+1-\sqrt n=\frac{2a+1-b}{a+1+\sqrt n}\geq\frac{a+1}{a+1+\sqrt n} > \frac a{a+\sqrt n}\geq\frac b{a+\sqrt n}=\sqrt n-a$$

• If $b\geq a+1$ then $(\sqrt n)=a+1$ because $$a+1-\sqrt n=\frac{2a+1-b}{a+1+\sqrt n}\leq\frac{a}{a+1+\sqrt n}<\frac{a+1}{a+\sqrt n}\leq\frac b{a+\sqrt n}=\sqrt n-a$$

Conclusion

• If $b\leq a$ then $n+(\sqrt n)=a^2+a+b$ and $$a^2< a^2+a+b< a^2+2a+1$$
• If $b\geq a+1$ then $n+(\sqrt n)=a^2+a+b+1$ and $$a^2+2a+1< a^2+a+b+1< a^2+4a+4$$

Answer 2

I'll reword your question, and I think it would still work. Let us take the square-rooted number and let that be $x$. Then, the possible values for the larger number $n$ is $x^2 - x \lt n \le x^2 + x$. So, it suffices to prove that there is no case where $x+n = k^2$.

Representing $n = x^2 + a$ where $-x \lt a \le x$, we get $x + x^2 + a = k^2$.

From here, we can get that $(k+x)(k-x) = x+a$.

Obviously, $k \gt x$, so the LHS is positive. Furthermore, $(k+x)(k-x) \gt x$, for obvious reasons. Therefore, we can eliminate all values of $ a \lt 0$.

So now, let us find the lower bound for $(k+x)(k-x)$. $k-x$, at its lowest, can be $1$. $k+x$, in this case, is $2x + 1$.

Note how the upper bound for the RHS was $a=x \to x+x = 2x$.

Therefore, it is impossible for this situation to be reached.