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Difference between revisions of "1986 IMO Problems/Problem 1"

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<math>d\equiv 2\pmod{4}: 2d-1</math> is not a perfect square.
 
<math>d\equiv 2\pmod{4}: 2d-1</math> is not a perfect square.
  
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Therefore, <math>d\equiv 1,–1 \pmod{4}.</math> Now consider <math>d\pmod{16}.</math>  
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Therefore, <math>d\equiv 1, \pmod{4}.</math> Now consider <math>d\pmod{16}.</math>  
  
 
<math>d\equiv 1,13 \pmod{16}: 13d-1</math> is not a perfect square.
 
<math>d\equiv 1,13 \pmod{16}: 13d-1</math> is not a perfect square.

Revision as of 10:18, 9 March 2018

Problem

Let $d$ be any positive integer not equal to $2, 5$ or $13$. Show that one can find distinct $a,b$ in the set $\{2,5,13,d\}$ such that $ab-1$ is not a perfect square.

Solution

We do casework with mods.

$d\equiv 0,3 \pmod{4}: 13d-1$ is not a perfect square.

$d\equiv 2\pmod{4}: 2d-1$ is not a perfect square.

Therefore, $d\equiv 1, \pmod{4}.$ Now consider $d\pmod{16}.$

$d\equiv 1,13 \pmod{16}: 13d-1$ is not a perfect square.

$d\equiv 5,9\pmod{16}: 5d-1$ is not a perfect square.

As we have covered all possible cases, we are done.

1986 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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