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Difference between revisions of "1986 IMO Problems/Problem 1"
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===Solution 1===
 
===Solution 1===
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We do casework with mods.
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We do casework with modular arithmetic.
  
 
<math>d\equiv 0,3 \pmod{4}: 13d-1</math> is not a perfect square.
 
<math>d\equiv 0,3 \pmod{4}: 13d-1</math> is not a perfect square.
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As we have covered all possible cases, we are done.
 
As we have covered all possible cases, we are done.
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~Shen kislay kai
  
 
===Solution 2===
 
===Solution 2===
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Suppose <math>p^2=2d-1</math>, <math>q^2=5d-1</math> and <math>r^2=13d-1</math>. From the first equation, <math>p</math> is an odd integer. Let <math>p=2k-1</math>. We have <math>d=2k^2-2k+1</math>, which is an odd integer. Then <math>q^2</math> and <math>r^2</math> must be even integers, denoted by <math>4n^2</math> and <math>4m^2</math> respectively, and thus <math>r^2-q^2=4m^2-4n^2=8d</math>, from which
 
Suppose <math>p^2=2d-1</math>, <math>q^2=5d-1</math> and <math>r^2=13d-1</math>. From the first equation, <math>p</math> is an odd integer. Let <math>p=2k-1</math>. We have <math>d=2k^2-2k+1</math>, which is an odd integer. Then <math>q^2</math> and <math>r^2</math> must be even integers, denoted by <math>4n^2</math> and <math>4m^2</math> respectively, and thus <math>r^2-q^2=4m^2-4n^2=8d</math>, from which
−
<cmath>2d=m^2-n^2=(m+n)(m-n)</cmath> can be deduced. Since <math>m^2-n^2</math> is even, <math>m</math> and <math>n</math> must either be odd or even simultaneously, so <math>(m+n)(m-n)</math> is divisible by <math>4</math>. It follows that the odd integer <math>d</math> must be divisible by <math>2</math>, leading to a contradiction. We are done.
+
<cmath>2d=m^2-n^2=(m+n)(m-n)</cmath> can be deduced. Since <math>m^2-n^2</math> is even, <math>m</math> and <math>n</math> have the same parity, so <math>(m+n)(m-n)</math> is divisible by <math>4</math>. It follows that the odd integer <math>d</math> must be divisible by <math>2</math>, leading to a contradiction. We are done.
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{{alternate solutions}}
 
{{alternate solutions}}
  
 
{{IMO box|year=1986|before=First Problem|num-a=2}}
 
{{IMO box|year=1986|before=First Problem|num-a=2}}

Latest revision as of 12:10, 3 September 2024

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

Solution 1

We do casework with modular arithmetic.

$d\equiv 0,3 \pmod{4}: 13d-1$ 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. ~Shen kislay kai

Solution 2

Proof by contradiction:

Suppose $p^2=2d-1$, $q^2=5d-1$ and $r^2=13d-1$. From the first equation, $p$ is an odd integer. Let $p=2k-1$. We have $d=2k^2-2k+1$, which is an odd integer. Then $q^2$ and $r^2$ must be even integers, denoted by $4n^2$ and $4m^2$ respectively, and thus $r^2-q^2=4m^2-4n^2=8d$, from which \[2d=m^2-n^2=(m+n)(m-n)\] can be deduced. Since $m^2-n^2$ is even, $m$ and $n$ have the same parity, so $(m+n)(m-n)$ is divisible by $4$. It follows that the odd integer $d$ must be divisible by $2$, leading to a contradiction. We are done.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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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