If a^n ≠ 0 and n is a positive integer, is n odd?

If a^n ≠ 0 and n is a positive integer, is n odd?

a^n ≠ 0 and n is a positive integer implies that a ≠ 0.

(1) a^n + a^(n+1) < 0 --> \((a+1)a^n<0\). Two cases:

\(a+1>0\) and \(a^n<0\). From first inequality we have that \(a>-1\). If a is a negative number (say -1/2), then \((negative)^n\) to be negative n must be odd.

\(a+1<0\) and \(a^n>0\). From first inequality we have that \(a<-1\), so a is a negative number. Now, \((negative)^n\) to be positive n must be even.

Not sufficient.

(2) a is an integer. Clearly insufficient.

(1)+(2) Consider the same two cases but now take into account that a is an integer:

\(a+1>0\) and \(a^n<0\). From first inequality we have that \(a>-1\). Since we know that \(a\neq{0}\), then a must be a positive integer (1, 2, 3, ...). Next, \((positive)^n\) cannot be negative, thus this case is out.

\(a+1<0\) and \(a^n>0\). From first inequality we have that \(a<-1\), so a is a negative integer. Now, \((negative)^n\) to be positive n must be even.

Sufficient.

Answer: C.

Hope it's clear.

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