Consider an infinite arithmetic progression of natural numbers. (a) Prove that if this progression contains cubes of integers then it contains infinitely

1.4.4. Consider an infinite arithmetic progression of natural numbers. (a) Prove that if this progression contains cubes of integers then it contains infinitely

many such terms. (b) Give an example of an infinite arithmetic progression of positive integers such that no term is the cube of an integer.

Solution. (a) Let (a n ) n ≥1

be an infinite arithmetic progression of natural numbers whose common difference is d. The idea is to show that if a k =q 3 1 is a cube, then (q 1 + md) 3 , with m ≥ 1, are all cubes in the arithmetic progression. Indeed, assume that a k =q 3 1 , with q 1 ∈ N. We find q 2 ∈ N such that there exists m 1 ∈ N satisfying

a +m d =q 3 3 +m d =q 3 2 k 2 1 2 ⇐⇒ q 1 1 2 ⇐⇒ m 1 d = (q 2 −q 1 )(q 1 +q 1 q 2 +q 2 ). Set q

2 =q 1 + d. We observe that in this case m 1 ∈ N, that is, a k +m 1 = (q 1 + d) . With the same argument for a k +m 1 we find a term a k +m 2 , with m 2 >m 1 , that is also a cube,

and so on. This shows that in fact, there are infinitely many terms with this property. (b) Let a be an odd positive integer. We show that the arithmetic progression defined

by a 1 = 2a and d = 4a 2 does not contain cubes of integers. Indeed, if 2a + 4a 2 k =q 3 , then 2a + 4a 2 k = 8q 3 2 1 3 . Hence a + 2a k = 4q 1 , impossible because the left-hand side is odd. ⊓ ⊔

An easier example consists in choosing an arithmetic progression with all mem- bers congruent to 2 (mod 7).

1.4.5. Let p k

be the k th prime number. Define the sequence (u n ) n ≥1 by u n =

p 1 + ··· + p n . Prove that between u n and u n +1 there is at least one perfect square.

Solution. We prove that if m 2 ≤u n then (m + 1) 2 −m 2 <u n +1 −u n =p n +1 . Indeed, observing that

2 u n + 1 ≥ 2m + 1 = (m + 1) 2 −m 2 , we have only to show that p

√ n +1 >2 u n + 1. This inequality may be rewritten as

The above inequality is true for n = 4 and then it is justified by induction, using p n +2 ≥p n +1 + 1. Let us now assume that there exists n ∈ N such that between u n

and u n +1 there is no perfect square. This means that there exists m ∈ N such that m 2 ≤u n and (m + 1) 2 ≥u n +1 . It follows that (m + 1) 2 −m 2 ≥u n +1 −u n , which contradicts the above relation. ⊓ ⊔

An interesting representation formula is proved in the next problem: any real number is a linear combination with integer coefficients of the terms of a given sequence converging to zero! It is also argued that this last assumption is sharp.

1.4 Qualitative Results 35

be a sequence of real numbers converging to 0 and contain- ing infinitely many nonzero terms. Prove that for any real number x there exists a sequence of integers ( λ

1.4.6. Let (a n ) n ≥1

n (x)) n ≥1 such that x = lim n → ∞ ∑ k =1 λ k (x)a k .

Is the assumption lim n → ∞ a n =0 necessary?

Solution. We can assume that a n x . We choose λ n (x) such that λ n (x)a n ≥ 0, for all n ∈ N. We first observe that there exist n 1 ∈ N and λ n 1 (x) ∈ Z such that λ n 1 (x)a n 1 ≤ x and |x− λ n 1 (x)a n 1 | < |a n 1 |. We point out that the existence of n 1 and λ n 1 (x) follows from lim n → ∞ a n = 0. For the same reasons, there exist n 2 ∈ N, n 2 >n 1 , and λ n 2 (x) ∈ Z such that λ n 2 (x)a n 2 ≤x− λ n 1 (x)a n 1 and |x − λ n 1 (x)a n 1 − λ n 2 (x)a n 2 | < |a n 2 |. Thus,

we obtain an increasing sequence of positive integers (n k ) and a sequence of integers ( λ n k (x)) n k such that

λ n k +1 (x)a n k +1 ≤x− λ n 1 (x)a n 1 − ··· − λ n k (x)a n k

and |x − λ n 1 (x)a n 1 − ··· − λ n k +1 (x)a n k +1 | < |a n k +1 |.

If n

1 ,n 2 ,...,n k , . . .} then we take λ n (x) = 0. Since lim k → ∞ a n k = 0, we obtain x = lim n → ∞ ∑ n k =1 λ k (x)a k . If x = 0, then we take λ n (0) = 0, and if x < 0, then we choose λ n (x) = − λ n (−x), for all n. If the sequence (a n ) does not converge to 0, then the result is not true. Indeed, it suffices to take a n = 1, for all n ≥ 1. If λ n ∈ Z and if the sequence defined by ∑ n k =1 λ k (x)a k is convergent, then its limit cannot belong ro R \ Z. ⊓ ⊔

The following is an interesting application of sequences in a problem of interest for many people. The statement is in connection with (easy!) IMO-type problems.

1.4.7. A chess player plays at least one game every day and at most 12 games every week. Prove that there exists a sequence of consecutive days in which he plays exactly 20 games.

be the number of games played in the first n days. By hyp- othesis, 21 ≤a 21 ≤ 36. Among the numbers a 1 ,a 2 ,...,a 21 there exist a i and a j (1 ≤ i < j ≤ 21) giving the same remainder on division by 20. Hence

Solution. Let a n

1 ≤a j −a i ≤ 35,

and 20 divides a j −a i . It follows that a j −a i = 20. Consequently, in the days

i + 1, i + 2, . . ., j, he played exactly 20 games. ⊓ ⊔ The method used for proving the next problem can be easily extended to deduce

similar properties of positive integers. Try to formulate some related properties!