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Re: Is k the square of an integer?
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19 Dec 2012, 02:05
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Expert Reply
Is k the square of an integer?
We need to determine whether k is a perfect square.
(1) k = t^2*q^6*r^10, where t, q, and r are integers --> \(k=(tq^3r^5)^2=integer^2\). Sufficient.
(2) k = t^2*m^15, where t and m are integers --> if m is a perfect square itself (say if m=n^2 for some integer n), then k will also be a perfect square but if m is NOT a perfect square, then k also will NOT be a perfect square. For example consider m=1 and m=2. Not sufficient.
Re: Is k the square of an integer?
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07 Feb 2022, 21:25
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A square of a number is derived when the number is multiplied by itself-> it has 2 copies of a number in a product
For K to be a square of an integer, it has to be an integer^2
1. k=t^2∗q^6∗r^10 where t, q and r are integers
sqrt(k) = sqrt(t^2∗q^6∗r^10) = t*q^3*r^5-> hence sqrt(K) is an integer -> sufficient
2.k=t^2*m^15 , where t and m are integers sqrt(k)=sqrt(t^2*m^15) = t*m^7*sqrt(m)-> now in this case we cannot say whether k is a square of an integer , if m=3, then K is not the square of an integer, if m=4 then K is a square of an integer-> hence insufficient
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