Are there two numbers, x and y, such that neither is a perfect square, and the difference of their roots is an integer? Can you find x and y such that the difference gets arbitrarily close? What about other powers?
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How about $x=y=2$? Then $\sqrt{x}-\sqrt{y}=0\in \mathbb{Z}$. This should work for any positive integer roots of 2. Or are you looking for distinct numbers? – SlipEternal Jul 23 '18 at 14:23
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Are we assuming that $x$ and $y$ are integers? If not, then there are infinite solutions, for example $3$ and $4+2\sqrt3$. – RayDansh Jul 23 '18 at 14:25
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Hint: Square both sides of the equation $\sqrt x=k+\sqrt y$. – Barry Cipra Jul 23 '18 at 14:30
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If I have understood the question, then what you ask is not possible.
Note that the the square root of an integer is either an integer itself or it is irrational. (this follows from the Rational Root Theorem)
If we have two distinct natural numbers $n,m$ such that $$\sqrt n - \sqrt m=a\in \mathbb Z$$ then $$\sqrt n=a+\sqrt m\implies n=a^2+2a\sqrt m+m\implies \sqrt m=\frac {n-a^2-m}{2a}$$
(Note: $a\neq 0$ since otherwise we'd have $n=m$.)
But this shows that $\sqrt m\in \mathbb Q$, hence it must be an integer, contradicting our assumptions.
lulu
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You can find distinct positive nonsquare integers whose difference is arbitrarily close to an integer, using the pigeonhole principle.
For instance if you take $100$ nonsquare positive integers, and look at their square roots $\pmod{1}$, two of these will have to be within $\frac{1}{100}$ of each other.
paw88789
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