Difference between revisions of "1969 IMO Problems/Problem 1"
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Both factors are positive, because if the left one is negative, then the right one would also negative, which is clearly false. | Both factors are positive, because if the left one is negative, then the right one would also negative, which is clearly false. | ||
− | It is also simple to prove that <math>n^2 + 2k^2 - 2nk > 1</math> when <math>k > 1</math>. Thus, for all <math>k > | + | It is also simple to prove that <math>n^2 + 2k^2 - 2nk > 1</math> when <math>k > 1</math>. Thus, for all <math>k > 1</math>, <math>4k^4</math> is a valid value of <math>a</math>, completing the proof. <math>\square</math> |
~mathboy100 | ~mathboy100 |
Latest revision as of 18:44, 10 August 2024
Problem
Prove that there are infinitely many natural numbers with the following property: the number is not prime for any natural number .
Solution
Suppose that for some . We will prove that satisfies the property outlined above.
The polynomial can be factored as follows:
Both factors are positive, because if the left one is negative, then the right one would also negative, which is clearly false.
It is also simple to prove that when . Thus, for all , is a valid value of , completing the proof.
~mathboy100
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1969 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |