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 > 2</math>, <math>4k^4</math> is a valid value of <math>a</math>, completing the proof. <math>\square</math>
+
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 $a$ with the following property: the number $z = n^4 + a$ is not prime for any natural number $n$.

Solution

Suppose that $a = 4k^4$ for some $k$. We will prove that $a$ satisfies the property outlined above.

The polynomial $n^4 + 4k^4$ can be factored as follows:

\[n^4 + 4k^4\] \[= n^4 + 4n^2k^2 + 4k^4 - 4n^2k^2\] \[= (n^2 + 2k^2)^2 - (2nk)^2\] \[= (n^2 + 2k^2 - 2nk)(n^2 + 2k^2 + 2nk)\]

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 $n^2 + 2k^2 - 2nk > 1$ when $k > 1$. Thus, for all $k > 1$, $4k^4$ is a valid value of $a$, completing the proof. $\square$

~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