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

Revision as of 14:02, 31 July 2022

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.

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

@@ Line 5: / Line 5: @@
 == Solution ==
-We do casework with mods.
+===Solution 1===
+We do casework with modular arithmetic.
-<math>d\equiv 0 \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.
 <math>d\equiv 2\pmod{4}: 2d-1</math> is not a perfect square.
-<math>d\equiv 3 \pmod{4}: 13d-1</math> is not a perfect square.
+Therefore, <math>d\equiv 1, \pmod{4}.</math> Now consider <math>d\pmod{16}.</math>
-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.
@@ Line 20: / Line 19: @@
 As we have covered all possible cases, we are done.
+===Solution 2===
+Proof by contradiction:
+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> 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.
+{{alternate solutions}}
 {{IMO box|year=1986|before=First Problem|num-a=2}}

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