I notice that the number seems to be a multiple of 3:
for n=2: $2^4 -1 = 15 $
for n=3: $2^6 -1 = 63$
for n=4: $2^8 -1 = 255$
How do I generalise?
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8It is quite difficult for a difference of squares to be a prime number. $a^2-b^2=(a-b)(a+b).$ – Jack D'Aurizio Dec 23 '15 at 14:11
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I feel really stupid now. Thanks you. – Gloomy Dec 23 '15 at 14:12
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2Possible duplicate of Why $a^n - b^n$ is divisible by $a-b$? – lab bhattacharjee Dec 23 '15 at 14:18
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This may be a specific case of a broader question, so to call it a duplicate is misleading, in my opinion. – Robert Soupe Dec 23 '15 at 17:16
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$2^{2n}-1=(2^n-1)(2^n+1)$, so it's always composite (unless $n=1$, in which case we have $3=(1)(3)$).
In addition, it's always a multiple of $3$, for the following reason. Since $2^n$ is never a multiple of $3$, it's either one more or one less than a multiple of $3$. This means that one of $2^n-1$ and $2^n+1$ is a multiple of $3$. Thus, $2^{2n}-1=(2^n-1)(2^n+1)$ is a multiple of $3$.
Akiva Weinberger
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If you have studied the $3x + 1$ problem even briefly, you know that the powers of 2 alternate between $-1$ and $1 \pmod 3$: $$2 = 3 - 1$$ $$4 = 3 + 1$$ $$8 = 9 - 1$$ $$16 = 15 + 1$$ $$32 = 33 - 1$$ This means that $2^{2n + 1} \equiv -1 \pmod 3$ and $2^{2n} \equiv 1 \pmod 3$. From this you just make a tiny hop to arrive at the conclusion that $2^{2n} - 1 \equiv 0 \pmod 3$. The only positive multiple of 3 that is prime is 3 itself. Therefore all $n > 1$ give composite multiples of 3.
Robert Soupe
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