1986 IMO Problems/Problem 1

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

 $5d-1=y^2\cdots (2)$

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

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 $x=3k$±$1$, $\implies 3|z$,

$\implies 9|z^2$.

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
Preceded by
First Problem
1 2 3 4 5 6 Followed by
Problem 2
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