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$\sqrt a$ is either an integer or an irrational number.

$\sqrt{2}$ is irrational number, but $\sqrt{9} = 3$ is an integer. Are there such integers whose square root is a (non-integer)rational number?

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    Well, integers are rational numbers. You presumably mean non-integer rational numbers.2012-09-05
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    Every integer is a rational number. I suppose you mean "Are there integers whose square root is rational, but not an integer?" In fact, the answer is no.2012-09-05
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    This is a duplicate of many prior questions, e.g. http://math.stackexchange.com/q/4467/242 and http://math.stackexchange.com/q/5/242 and http://math.stackexchange.com/q/26499/242 and http://math.stackexchange.com/q/22423/242 and http://math.stackexchange.com/q/11872/2422012-09-05
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    @GeoffRobinson (you should) write it as an answer.2012-09-05
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    @tzador, please read the duplicate question. It says that the square root of every (positive) integer is either an integer or irrational. There is no integer whose square root is a rational fraction.2012-09-05
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    One way of looking at this is to try taking the square of a rational number $q$ which is not an integer. If this is in its lowest terms, then the denominator of $q$ will have a non-trivial prime factor $p$ and $q^2$ will have denominator divisible by $p^2$. I mention this because the ideas here, though seemingly trivial, have fruitful and interesting generalisations.2012-09-05
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    @Jennifer: It is a duplicate, so no real need.2012-09-05

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