Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $k,l,m,n,p\in\mathbb N$ and a prime number $p>3$ that satisfies $$p^k+p^l+p^m=n^2$$ is chosen, prove that $8\mid p+1$.

$n^2$, when divided by $8$, gives a remainder $1$ (it can't give the remainders $0$ and $4$, because three odd numbers sum up to an odd number, which, if it is a square, always gives a remainder 1 when divided by 8).

The prime numbers $p$ give these ones:

When $p\equiv 1\pmod 8$, then $p^{2z}\equiv 1\pmod 8$ and $p^{2z+1}\equiv 1\pmod 8$.

When $p\equiv 3\pmod 8$, then $p^{2z}\equiv 1\pmod 8$ and $p^{2z+1}\equiv 3\pmod 8$.

When $p\equiv 5\pmod 8$, then $p^{2z}\equiv 1\pmod 8$ and $p^{2z+1}\equiv 5\pmod 8$.

When $p\equiv 7\pmod 8$, then $p^{2z}\equiv 1\pmod 8$ and $p^{2z+1}\equiv 7\pmod 8$.

Where $z\in\mathbb N\cup \{0\}$. We have to prove that $p\equiv 7\pmod 8$.

And another observation is that we could mark $p$ as either $3c+1$ or $3c+2$, where $c\in\mathbb N\cup\{0\}$ (we have to use the fact that $p>3$ somehow anyway). Thanks.

And I've given a tag "diophantine-equations" to this question because this seems a bit related to them. Feel free to disagree.

share|improve this question
    
$n^2\equiv 1\pmod 8$ because the sum of three odd numbers is odd. –  Ian Mateus Dec 28 '13 at 0:08
    
Oh, good point. I've edited the question. –  mathh Dec 28 '13 at 0:10
2  
the significance of 3 is only that 3+3+3=9. which highlights the fact that the exponents are allowed to be 1. you need to deal with that...and 27+27+27 = 81. So, the real extra case is the three exponents equal.. –  Will Jagy Dec 28 '13 at 0:13
    
@WillJagy The significance is actually that $8\nmid 3+1$. –  mathh Feb 12 at 13:25

1 Answer 1

up vote 4 down vote accepted

When $p\equiv 1\mod 8$, we now that $P=p^k+p^l+p^m\equiv 1+1+1\equiv 3\not \equiv 1\mod 8$. When $p\equiv 5\mod 8$, we get $P\equiv 1+1+1\equiv 3\not\equiv 1\mod4$, so this won't work either. When $p\equiv 3\mod 8$, we get $3+3+3\equiv 1\mod 8$ (and the other possibilities won't work), so the only case we still have to shoot is $k\equiv l\equiv m\equiv 1\mod 2$ with $p\equiv 3\mod 8$.

Now, we know $p|n$, so $p^2|n^2$, so $p^2|P$, so $k,l,m\geq 2$ (because $p>3$). Thus $k,l,m\geq 3$, because they are odd. Now, we get $p^3|n^2$, so $p^2|n$, so $p^4|n^2$. Thus, $k,l,m\geq 5$. This will continue forever, so there aren't any solutions in this case. The only remaining case is what you have to prove, so we are done.

share|improve this answer
    
When $p\equiv5\pmod 8$, then $p^k,p^l,p^m\equiv 1$ or $5\pmod 8$, because we don't know whether $k,l,m$ are even or odd. –  mathh Dec 28 '13 at 0:38
    
But I am working $\mod 4$ in that equation. and $1\equiv 5\mod 4$. –  Ragnar Dec 28 '13 at 0:39
    
Oh, hadn't noticed. –  mathh Dec 28 '13 at 0:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.