# interesting one

Find all primes p and q ,and even numbers n > 2 , satisfying the equation
pn + pn-1 + Â· Â· Â· + p + 1 = q2 + q + 1.

11
Devil ·

If pâ‰ q, then we have (p-1)+(p-1)a=2q, for some integer a, that gives p-1=2 or p-1=q, excluding the case when p=2. That gives p>q, implying nâ‰¤2

21
Shubhodip ·

pls write a detailed proof

21
Shubhodip ·

ahh! at last i got it!

with n= 2k, k>1 it is

p(pk+ 1)(pk-1+ ... + p+ 1)= q(q+1)>p(p+1)>30

so q>2

gcd((pk+ 1),Pk-1+ ... + 1) = gcd(pk+1, pk-1p-1)â‰¤ gcd(pk+1, pk-1)â‰¤2

easy to see that p|q+1 (*)

Case 1: q|pk-1+ ...+ 1,(then qâ‰¤ pk-1+ ...+ 1) ,so q doesn't divide pk+ 1, i.e gcd(q,pk+1)= 1

easy to see that pk+1|q+1, implying pk+1â‰¤q+1

since gcd(p,pk+1)= 1, p(pk+1)|q+1 from (*)

so p(pk+1)â‰¤q+1â‰¤pk-1+ ...+ p+2

which is wrong since pk+1 -1 + p> pk-1> (pk-1p-1)

Case 2: q| pk+1,(-) so q does not divide pk-1+ ... + p+ 1 i.e gcd(q,pk-1+ ... + p+ 1)= 1

easy to see that pk-1+ ... + p+ 1|q+1

agin gcd(p,pk-1+ ... + p+ 1)= 1

so p(pk-1+ ... + p+ 1)|q+1

it follows that p(pk-1+ ... + p+ 1)â‰¤q+1â‰¤pk+2, from (-)

which means pk+ ...+ pâ‰¤pk+ 2

i.e p(pk-2+ ...+ 1)â‰¤2

only possibility p=2,k= 2 giving q= 5

so the only solution is p=2,q=5,n= 4, and its easy to check that this indeed is a solution..

PS. @devil: i dont understand your hint, can you pls put details to it?

@h4hemag: hey is this a real olympiad question ?