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

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Joined: 02 Sep 2009
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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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21 Mar 2018, 05:31
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Difficulty:

45% (medium)

Question Stats:

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

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If m and n are both positive integers is $$\sqrt[6]{m + n^4}$$ an integer?

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

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

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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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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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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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