1975 IMO Problems/Problem 2
Problem
Let be an infinite increasing sequence of positive integers. Prove that for every there are infinitely many which can be written in the formwith positive integers and .
Solution
If we can find such that , we're done: every sufficiently large positive integer can be written in the form . We can thus assume there are no two such . We now prove the assertion by induction on the first term of the sequence, . The base step is basically proven, since if we can take and any we want. There must be a prime divisor which divides infinitely many terms of the sequence, which form some subsequence . Now apply the induction hypothesis to the sequence .
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]
See Also
1975 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
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This is flawed. At Least it is not clearly shown how a sequence with infinite number of terms that are coprime to a(p) is achieved by simply repeating the procedure as described. Following proof would be much more convincing. For every p>=1 we calculate the remainder of each term of the sequence modulo a(p) and we will have a(p) types of remainders: 1~a(p). There are infinitely many terms and limited types of remainders therefore, there must be one remainder which belongs to infinitely many terms. Let it be r, 1<=r<=a(p), the first term of such remainder be a(q) and the subsequent terms be a(m), m towards infinity. Obviously a(m)>a(q)>=a(p)+1, and m>q>p. Pick any a(m)>a(q), we now prove that it can always be written in the form a(m)=xa(p)+ya(q). Let a(q)=k1•a(p)+r, a(m)=k2•a(p)+r, k2>k1. Then a(m)-a(q)=(k2-k1)a(p), a(m)=(k2-k1)a(p)+1•a(q). Let x=k2-k1, y=1, and obviously both are positive integers so we have achieved the target form. As there are infinitely many a(m), for every p>=1, we will always have infinitely many a(m) which can be written in the target form. Done.