Difference between revisions of "1969 IMO Problems/Problem 1"

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.

@@ Line 14: / Line 14: @@
 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

1969 IMO (Problems) • Resources
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Difference between revisions of "1969 IMO Problems/Problem 1"

Latest revision as of 18:44, 10 August 2024

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