Difference between revisions of "1969 IMO Problems/Problem 1"
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==Solution== | ==Solution== | ||
− | + | Suppose that <math>a = 4k^4</math> for some <math>a</math>. We will prove that <math>a</math> satisfies the property outlined above. | |
+ | |||
+ | The polynomial <math>n^4 + 4k^4</math> can be factored as follows: | ||
+ | |||
+ | <cmath>n^4 + 4k^4</cmath> | ||
+ | <cmath> = n^4 + 4n^2k^2 + 4k^4 - 4n^2k^2</cmath> | ||
+ | <cmath> = (n^2 + 2k^2)^2 - (2nk)^2</cmath> | ||
+ | <cmath> = (n^2 + 2k^2 - 2nk)(n^2 + 2k^2 + 2nk)</cmath> | ||
+ | |||
+ | 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 > 2</math>, <math>4k^4</math> is a valid value of <math>a</math>, completing the proof. <math>\square</math> | ||
+ | |||
+ | ~mathboy100 | ||
{{alternate solutions}} | {{alternate solutions}} | ||
== See Also == {{IMO box|year=1969|before=First question|num-a=2}} | == See Also == {{IMO box|year=1969|before=First question|num-a=2}} |
Revision as of 18:13, 7 December 2022
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 |