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A Woodal number is an integer of the form $n 2^{n}-1$.

A Woodal prime is an integer that is both a prime and a Woodal number.

Let $p$ be a prime of the form 1 mod 4.

Then $p 2^{p} -1$ is never a ( Woodal ) prime.

How to prove this ?

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Just a try - p is of form 1 mod 4, now let's look on 2^p. when p>2, 2^p will always be 0 mod 4 (because 4 is 2^2 and by rules of power it divides with other powers of 2 without remainder). So if we multiply p with 2^p the result will be 0 mod 4. in addition, (-1) mod 4 is 3, so if you add both numbers, p*2^p - 1 mod 4 would be 3. meaning, it's divided by 4 with remainder of 3, but divided with 3 without remainder. which means it can't be a prime because 3 is one of his factors –  DanielY May 30 '13 at 7:49
    
Take p=13. 13 = 1 mod 4. Now (13 2^13 - 1)/5 is an integer but it is not a multiple of 3. So your argument fails ?? –  mick Jun 1 '13 at 21:17
    
Consider only 13*2^13 - 1, without /5. Because this problem asks about woodal primes in this form –  DanielY Jun 2 '13 at 10:56
    
never mind. you're right –  DanielY Jun 2 '13 at 16:55

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