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 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.

Solution 3

By contradiction suppose \begin{align} 2d-1 &= a^2 \\ 5d-1 &= b^2 \\ 13d-1 &= c^2 \end{align} Note $n^2 \equiv 1 \text{ or } 0 \mod 4$ . By the first equation $d = 2d_1 + 1$ otherwise $-1 \equiv a^2 \mod 4$ . Substituting, the three equations become \begin{align} 4d_1 + 1 &= a^2 \\ 10d_1 + 4 &= b^2 \\ 26d_1 + 12 &=c^2 \end{align} By the second equation $2|b$ hence $4|b^2$ hence $2|d_1$ . Write $d_1=2d_2$ . Substituting, the three equations become \begin{align} 8d_2 + 1 &= a^2 \\ 5d_2+1 &= b_1^2 \\ 13d_2 + 3 &= c_1^2 \end{align} where $b = 2b_1, c = 2c_1$ . Taking mod $3$ of the third equation implies $d_2 \equiv c_1^2 \equiv 0 \text{ or } 1 \mod 3$ . Case $d_2 = 3d_3 + 1$ : substituting, the first equation becomes $24d_3 + 9 = a^2$ . Similar to before this implies $3|d_3$ . The second equation becomes $15d_3 + 6 = b_1^2$ . Since $3 | b_1$ and $9|b_1^2$ we get $9|6$ . A contradiction. Case $d_2 = 3d_3$ : the third equation becomes $13d_3+1 = 3c_2^2$ where $c_1 = 3 c_2$ . Taking mod $3$ yields $d_3 +1 \equiv 0 \text{ or } 1 \mod 3$ . The only way this is possible is if $d_3 = 3 d_4$ . But then $13\cdot 3 d_4 + 1 = 3c_2^2$ which implies $3 | 1$ . So we are done.

~detriti

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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