Question

# Let a, b, c be positive real numbers. The sequence (an)n≥1 is defined bya1=a,a2=b andan+1=a2n+can−1 for all n≥2. Prove that the terms of the sequence are all positive integers if and only if a,b and a2+b2+cab are positive integers.

A
c=asas+2a2s
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B
c=asas2a2s
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C
c=asas2+a2s
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D
c=asas+2+a2s
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Solution

## The correct option is A c=asas+2−a2sClearly, all the terms of the sequence are positive numbers. Write the recursive relation as an+1an−1=a2n+c.anan−2=a2n−1+c,and by subtracting the two equalities we deduce an−1(an+1+an−1)=an(an+an−2).Therefore an+1+an−1an=an+an−2an−1n≥3 for allbn=an+1+an−1an It follows that the sequencebn=k, is constant, sayFor all n≥2. Then the sequence (an) satisfies the recursive relation an+1=kan−an−1 for all n≥2 and since a3=b2+ca=kb−a, We derive that k=a2+b2+cab Now, if a,b and k are positive integers, it follows inductively that anis a positive integer for all n≥1Then a,b are positive integers and k=a3+ab is a rational number .Letk=pq,where p and q are relatively prime positive integers. We want to prove that q = 1. Suppose that q>1. From the recursive relation we obtain q(an+1+an−1)=pan, and hence q divides an for all n≥2 We prove by induction on s that qs divides an for all n≥s+1 We have seen that this is true for s=1. Suppose qs−1 divides an for all n≥s. we have an+2=pqan+1−an,which is equivalent toan+2qs−1=pan+1qs−anqs−1. if n≥s, then,qs−1 divides an hence an+2, and qs divides an+1. it follows that qsas+2=a2s+1+casc=asas+2−a2swhich implies that c is divisible by q2(s−1) for all s≥1. because c>0, this is a contradiction.

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