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The following puzzle is from the December 2012 situation of the Eureka journal (revealed by The Cambridge University Mathematical Society).
Find all integers n such that $sqrt{n-4 sqrt{n-19}}$ can be an integer.
To be accepted, your reply should present that you’ve got discovered all integers satisfying the given situation.
requested Jul 31 at 21:54
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10
If $sqrt{n-4sqrt{n-19}}$ is an integer, then $sqrt{n-19}$ should even be an integer, as a result of sq. roots of integers are both integer or irrational. Let us name this integer $m=sqrt{n-19}$. We rewrite the expression in phrases on $m$:
$$sqrt{m^2+19-4m}$$which have to be an integer. Therefore,
$$m^2+19-4m$$
have to be an ideal sq. or, equivalently,
$$(m-2)^2+15$$
have to be an ideal sq.. As $(m-2)^2$ can be an ideal sq., we should discover two excellent squares whose distinction is 15. As the variations between consecutive excellent squares enhance monotonically, it’s simple to see that there are solely two pairs of excellent squares that differ by 15: $(1, 16)$ and $(49, 64)$. The smaller sq. is $(m-2)^2$, so we now have 4 choices:
– $m-2=-1 implies m=1 implies n=20$.
– $m-2=1 implies m=3 implies n=28$.
– $m-2=-7 implies m=-5 implies n=44$. This answer is not legitimate.
– $m-2=7 implies m=9 implies n=100$.So there are three options: $n=20$, $n=28$, $n=100$. The worth $n=44$ isn’t legitimate as a result of it could require the unfavorable sq. root of $n-19$.
$endgroup$
