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 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.

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.

Solution 3

Suppose one can't find distinct a,b from the set $A=\{2,5,13,d\}$ such that $ab-1$ is a perfect square.

Let, $2d-1=x^2\cdots (1)$

$13d-1 =z^2 \cdots (3)$ .b

Clearly $Z^2+1 = 13d = 3(5d)-2d= 3y^2-x^2+2$ .

$\implies x^2 +z^2=3y^2+1$ .

Clearly ,if $x^2,z^2$ is 1 or 0 modulo 3 then it has no solution .

Suppose, $Z=3r$ and $x=3$ ± $1$ , $\implies 3|z$ ,

$\implies 9|z^2$ .

So, $5d-1 \equiv 0 \pmod{9}$ and $13d-1 \equiv 0 \pmod{9}$ .

$\implies d \equiv 0 \pmod{d}$ .

It is contradiction ! Since $9|5d-1$ .

@ftheftics

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
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1986 IMO Problems/Problem 1

Contents

Problem

Solution

Solution 1

Solution 2

Solution 3