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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The proof of this factorisation can be done by geometric series, first with common ratio \( p\), then with common ratio \(p^2\).
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. \(p+1 \ge 4\).
where all of the \(c_i \) are assumed to satisfy \( 0 \le c_i < 2^N \)
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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.)