Difference between revisions of "1969 IMO Problems/Problem 1"
(Created page with '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</…') |
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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 z is composite for all <math> a = 4 m^4</math> | + | ==Problem== |
| + | Prove that there are infinitely many natural numbers <math>a</math> with the following property: the number <math>z = n^4 + a</math> is not prime for any natural number <math>n</math>. | ||
| + | |||
| + | ==Solution== | ||
| + | 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> | ||
| + | |||
| + | {{alternate solutions}} | ||
| + | |||
| + | == See Also == {{IMO box|year=1969|before=First question|num-a=2}} | ||
Revision as of 12:35, 29 January 2021
Problem
Prove that there are infinitely many natural numbers 
with the following property: the number 
is not prime for any natural number 
.
Solution
The equation was ,you can put 
for all natural numbers m. So you will get 
so you get 
is composite for all
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 | ||
