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> | + | Suppose that <math>a = 4k^4</math> for some <math>k</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: | The polynomial <math>n^4 + 4k^4</math> can be factored as follows: | ||
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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:
![\[n^4 + 4k^4\]](http://latex.artofproblemsolving.com/7/2/3/72361522cb7d2c5daa5ae75eb77f74dbfdcece7f.png)
![\[= n^4 + 4n^2k^2 + 4k^4 - 4n^2k^2\]](http://latex.artofproblemsolving.com/3/5/7/357daa0e2be5a29f749aed9641348e55c19b9dd3.png)
![\[= (n^2 + 2k^2)^2 - (2nk)^2\]](http://latex.artofproblemsolving.com/3/f/4/3f43938370e83a32d5f64fa354b0313538bd146f.png)
![\[= (n^2 + 2k^2 - 2nk)(n^2 + 2k^2 + 2nk)\]](http://latex.artofproblemsolving.com/4/4/c/44cc5332167252a24cfbe3fdb6fe6c68ba95ac2e.png)
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 | ||
