p and since (c + n1)k· pk−1− c
is a non negative integer the perfect jkth power xjk belongs to the arithmetic progression

Problem 11182

(American Mathematical Monthly, Vol.112, November 2005) Proposed by S. Amrahov (Turkey).

Lethani be an arithmetic progression of positive integers for which the common difference is prime.

Given that the sequence includes both a term that is a perfect jth power and a term that is a perfect kth power, and that j and k are relatively prime, prove that there exists a term that is a perfect jkth power.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

The term of the arithmetic progression is given by an = a0+ n · p for some prime p and for n ∈ N.

By hypothesis there are n1, n₂∈ N such that

a₀+ n1· p = x^j and a0+ n2· p = y^k for some positive integers x and y.

If a0is a multiple of p, say a0= c · p then

x^jk= (a0+ n1· p)^k= (c + n1)^k· p^k−1
· p = a0+ (c + n1)^k· p^k−1− c
· p

and since (c + n1)^k· p^k−1− c
is a non negative integer the perfect jkth power x^jk belongs to the arithmetic progression.

Now assume that a0 is not a multiple of p. Since a0= x^j = y^k mod p then for s, t ∈ Z a^ks+jt₀ = a^ks₀ · a^jt₀ = x^jks· y^kjt = x^s· y^t
jk

mod p.

Since j and k are relatively prime there are some s0, t₀∈ Z such that ks0+ jt0 = 1. Moreover, by Euler-Fermat theorem a^p−1 = 1 mod p (a0 is not a multiple of p), hence letting s = s0+ a · (p − 1) and t = t0+ b · (p − 1) for a, b ∈ N we have that

x^s· y^t
jk

= a^ks+jt₀ = a1+(a+b)(p−1)

0 = a0 mod p.

Therefore there is an integer n3 such that

a₀+ n3· p = x^s· y^t
jk .

and by taking the integers a and b big enough we can be sure that s and t are non negative. Hence the perfect jkth power(x^s· y^t)^kj belongs to the arithmetic progression because it is greater than a0

and therefore n3is non negative.