Difference between revisions of "1975 IMO Problems/Problem 2"
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+ | If we can find <math>p\ne q</math> such that <math>(a_p,a_q)=1</math>, we're done: every sufficiently large positive integer <math>n</math> can be written in the form <math>xa_p+ya_q,\ x,y\in\mathbb N</math>. We can thus assume there are no two such <math>p\ne q</math>. We now prove the assertion by induction on the first term of the sequence, <math>a_1</math>. The base step is basically proven, since if <math>a_1=1</math> we can take <math>p=1</math> and any <math>q>1</math> we want. There must be a prime divisor <math>u|a_1</math> which divides infinitely many terms of the sequence, which form some subsequence <math>(a_{k_n})_{n\ge 1},\ k_1=1</math>. Now apply the induction hypothesis to the sequence <math>\left(\frac{a_{k_n}}u\right)_{n\ge 1}</math>. | ||
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+ | The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [https://aops.com/community/p367494] | ||
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+ | == See Also == {{IMO box|year=1975|num-b=1|num-a=3}} |
Latest revision as of 15:10, 29 January 2021
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 |
All IMO Problems and Solutions |