If p is a prime number and k is a positive integer that is greater than one. Prove 
cannot be expressed as a power of 2.
Can someone help me to prove this :O
noting, of course, \( k > 1\).
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\left(p^{2\ell} + p^{2\ell-2} + \dots + p^2 + 1\right))
The proof of this factorisation can be done by geometric series, first with common ratio \( p\), then with common ratio \(p^2\).
\text{ are factors of a power of 2}\\ \text{they must }\textbf{also both}\text{ be powers of 2}\\ \text{noting they are, obviously, integers.})
This is just the basic fact that if \(x = ab\) then \(a \) and \(b \) must be
factors of \(x\). Here, \(x = 2^m\) for some integer \(m\).
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Now for the handwavy bit.
Note: The restriction on \(N\) is enforced by the fact that \(p \ge 3\), i.e. \(p+1 \ge 4\).
}\\ \text{and index laws, we note that}\\ (2^N-1)^{2\ell} + (2^N-1)^{2\ell -2} + \dots + (2^N - 1)^2 + 1 = (\text{a bunch of terms of which }2^N\text{ must be a factor} + \ell + 1)\\ \text{To give those terms some coefficients, we rewrite this as}\\ c_{2N\ell} 2^{2N\ell} + \dots + c_{N + 1}2^{N+1} + c_{N} 2^{N}+ \ell + 1)
where all of the \(c_i \) are assumed to satisfy \( 0 \le c_i < 2^N \)
 causes the problem.})
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From here, a sketch will be given as the formal write-up can be messy.
In the first case, suppose that \(2^N\) is not a factor of \( \ell + 1\). Instead, suppose that \( 2^M\) is the maximum power on 2 that is a factor of \( \ell + 1\), where \(M < N\). Upon factoring out \(2^M\), equation \( (*) \) will be \(2^M\) times something.
That "something" will be the sum of a bunch of things divisible by \( 2^{N-M} \), plus something that is not (because in fact, isn't divisible by ANY power of 2). When you add something that is a power of 2 to something that is not even divisible by 2, you will get something that is not a power of 2.
Hence, the long winded expression cannot be a power of 2 either, which is the contradiction.
In the second case, where \(2^N\)
is a factor of \(\ell + 1\), we rinse and repeat but shift the focus from \( \ell + 1\) to \(c_N\). (Basically, the problematic term changes.)
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