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Difference between revisions of "1969 IMO Problems/Problem 1"
 
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==Solution==
 
==Solution==
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The equation was <math>z = n^4 + a</math> ,you can put <math>a = 4 m^4 </math> for all natural numbers m. So you will get <math> z = n^4 + 4 m^4 = n^4+4m^4 +4n^2 m^2 - 4n^2 m^2</math> <math>z = (n^2+2 m^2)^2 - (2nm)^2 = (n^2+2 m^2 -2nm)(n^2+2 m^2 +2nm) </math> so you get <math>z</math> is composite for all <math> a = 4 m^4</math>
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Suppose that <math>a = 4k^4</math> for some <math>a</math>. We will prove that <math>a</math> satisfies the property outlined above.
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The polynomial <math>n^4 + 4k^4</math> can be factored as follows:
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<cmath>n^4 + 4k^4</cmath>
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<cmath> = n^4 + 4n^2k^2 + 4k^4 - 4n^2k^2</cmath>
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<cmath> = (n^2 + 2k^2)^2 - (2nk)^2</cmath>
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<cmath> = (n^2 + 2k^2 - 2nk)(n^2 + 2k^2 + 2nk)</cmath>
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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.
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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>
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~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}}

Latest revision as of 18:13, 7 December 2022

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 $a$. 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 $k > 2$, $4k^4$ 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
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