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If m and n are both positive integers is (m + n^4)^(1/6) an integer?

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If m and n are both positive integers is (m + n^4)^(1/6) an integer?  [#permalink]

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New post 21 Mar 2018, 05:31
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  45% (medium)

Question Stats:

70% (01:45) correct 30% (01:52) wrong based on 54 sessions

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Re: If m and n are both positive integers is (m + n^4)^(1/6) an integer?  [#permalink]

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New post 21 Mar 2018, 06:34
Bunuel wrote:
If m and n are both positive integers is \(\sqrt[6]{m + n^4}\) an integer?


(1) \(m = n^4 + n^5\)

(2) \(m = n^4(n^2 - 1)\)


1)
n^4 + n^5 = m

\(\sqrt[6]{m + n^4}\) = \(\sqrt[6]{n^4 + n^5 + n^4}\) = \(\sqrt[6]{n^5 + 2n^4}\) = insufficient

2) \(m = n^4(n^2 - 1)\) = \(m = n^6 - n^4\)
\(\sqrt[6]{m + n^4}\) = \(\sqrt[6]{n^6 - n^4 + n^4}\) = n

(B) imo
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Re: If m and n are both positive integers is (m + n^4)^(1/6) an integer?  [#permalink]

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New post 05 Dec 2018, 10:00
Hatakekakashi wrote:
Bunuel wrote:
If m and n are both positive integers is \(\sqrt[6]{m + n^4}\) an integer?


(1) \(m = n^4 + n^5\)

(2) \(m = n^4(n^2 - 1)\)


1)
n^4 + n^5 = m

\(\sqrt[6]{m + n^4}\) = \(\sqrt[6]{n^4 + n^5 + n^4}\) = \(\sqrt[6]{n^5 + 2n^4}\) = insufficient

2) \(m = n^4(n^2 - 1)\) = \(m = n^6 - n^4\)
\(\sqrt[6]{m + n^4}\) = \(\sqrt[6]{n^6 - n^4 + n^4}\) = n

(B) imo



Just by looking at statement 1 , can anybody elaborate by taking some values of n, how it is insufficient?

Kudos for a good explanation! Thank you.
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Re: If m and n are both positive integers is (m + n^4)^(1/6) an integer?   [#permalink] 05 Dec 2018, 10:00
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