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How would I go about doing this?
If p is a prime number so that p + 1 is a cubed number, prove that p = 7
I'm at a loss.
I'm also at a loss on if p > q >= 5 are prime numbers, then prove that 24 divides (p^2-q^2)
please help!!
Thanks so much
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All primes are odd apart from 2, so p+1 either equals 2+1=3 or an (odd number +1)=even number "n^3".
As 3 is not a cubed integer we know that p≠2. This means "n^3" is even, thus meaning "n" must be even as (even)^3=even and (odd)^3=odd, and as 2 is the only even prime number, n=2
therefore, p+1=2^3=8
Meaning p=7
Last edited by adam_dsutton (2007-09-20 12:15:46)
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Adam, the question only says p+1 is a cubed number; it does not say that it must be the cube of a prime number!
This is how to prove it.
Last edited by JaneFairfax (2007-09-20 21:02:46)
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For the second part, write p[sup]2[/sup]−q[sup]2[/sup] = (p+q)(p−q)
Since p and q are both odd primes, p+q and p−q are both even. Also, one of them must be divisible by 4. Proof:
Either p−q is divisible by 4 or its not. If its divisible by 4, then thats it. If not, let p−q = 4k[sub]1[/sub]+2. Then p+q = p−q+2q = 4k[sub]1[/sub]+2(q+1); as q+1 is even, 2(q+1) is a multiple of 4; ∴ p+q is divisible by 4. QED.
Since p+q and p−q are both even and one of them is divisible by 4, their product is divisible by 8.
Finally, since p and q are both not divisible by 3, p = 3k[sub]2[/sub]±1 and q = 3k[sub]3[/sub]±1; going through all the possibilities, we also find that one of p+q and p−q is divisible by 3. (In general: if three numbers form an arithmetic progression whose common difference is not a multiple of 3, one of those three numbers must be divisible by 3. This is a useful result to bear in mind. (In this case, p−q, p, and p+q form an arithmetic progression.))
So (p+q)(p−q) is divisible by both 8 and 3; ∴ its divisible by 24.
Last edited by JaneFairfax (2007-09-20 21:12:20)
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