IMO 1986 Problem 1

IMO 1986 Problem 1

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 15m13s

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.

Exploration

Assume, for contradiction, that all six expressions

$2\cdot 5-1,;2\cdot 13-1,;5\cdot 13-1,;2d-1,;5d-1,;13d-1$

are perfect squares. The first three are already known to be squares:

$2\cdot 5-1=9,\quad 2\cdot 13-1=25,\quad 5\cdot 13-1=64.$

The remaining conditions force a system

$2d-1=x^2,\quad 5d-1=y^2,\quad 13d-1=z^2.$

Small checks show consistency in each pair separately:

for example, $d=1,3,5,7$ fail at least one equation immediately, and no small value suggests simultaneous solvability of all three equations. This suggests the obstruction must come from incompatibility between the three Diophantine constraints rather than from any single one.

The previous approach failed because it produced auxiliary congruences without proving they exclude all solutions. A correct argument must isolate a structural invariant that cannot simultaneously satisfy both derived Pell-type equations.

Problem Understanding

We must prove that for every positive integer $d\neq 2,5,13$, among the numbers

$2d,;5d,;13d,$

at least one of $ab-1$ for $a,b\in{2,5,13,d}$ is not a perfect square. Since

$2\cdot 5-1=9,\quad 2\cdot 13-1=25,\quad 5\cdot 13-1=64,$

the only potentially problematic pairs involve $d$, so it suffices to show that the system

$2d-1=x^2,\quad 5d-1=y^2,\quad 13d-1=z^2$

has no integer solution in positive integers $d$.

Key Observations

From

$2d-1=x^2$

we obtain

$d=\frac{x^2+1}{2},$

so $x$ is odd.

Substituting gives two conic equations:

$2y^2-5x^2=3,\qquad 2z^2-13x^2=11.$

Each equation defines an integer sequence of admissible $x$ values. The key structural point is that these two sequences of $x$ values lie in different residue classes modulo $4$, which forces incompatibility.

The required rigidity does not come from weak congruences like modulo $8$, but from a descent argument that pins down a fixed parity class for $x$ in each equation separately.

Solution

Lemma 1

All integer solutions of

$2y^2-5x^2=3$

satisfy $x\equiv 1 \pmod 4$.

Proof.

Assume $(x,y)$ is an integer solution. From

$2y^2=5x^2+3,$

we deduce $y$ is even, so write $y=2u$. Then

$8u^2=5x^2+3.$

Working modulo $8$, since $8u^2\equiv 0$, we obtain

$5x^2+3\equiv 0 \pmod 8.$

Because $5\equiv 5^{-1}\equiv 5 \pmod 8$, this yields

$x^2\equiv 1 \pmod 8,$

so $x$ is odd.

We now refine the structure. If $x\equiv 3 \pmod 4$, write $x=4k+3$. Then

$x^2=16k(k+1)+9,$

so

$5x^2+3=80k(k+1)+48=16(5k(k+1)+3).$

Thus $u^2=2(5k(k+1)+3)$ is even, so $u$ is even, say $u=2v$. Substituting gives

$32v^2=5x^2+3,$

hence

$16v^2=5k(k+1)+3.$

The left-hand side is divisible by $16$, while the right-hand side satisfies

$5k(k+1)\equiv 0 \text{ or } 5 \pmod{8},$

so $5k(k+1)+3\equiv 3 \text{ or } 0 \pmod 8$, never $0 \pmod{16}$. This contradiction excludes $x\equiv 3\pmod 4$.

Hence $x\equiv 1\pmod 4$. ∎

Lemma 2

All integer solutions of

$2z^2-13x^2=11$

satisfy $x\equiv 3 \pmod 4$.

Proof.

From

$2z^2=13x^2+11,$

we see $z$ is even, so write $z=2w$. Then

$8w^2=13x^2+11.$

Reducing modulo $8$ gives

$13x^2+11\equiv 0 \pmod 8.$

Since $13\equiv 5 \pmod 8$, this becomes

$5x^2+11\equiv 0 \pmod 8,$

hence

$5x^2\equiv 5 \pmod 8,$

so

$x^2\equiv 1 \pmod 8,$

and $x$ is odd.

Now assume $x\equiv 1 \pmod 4$, write $x=4k+1$. Then

$x^2=16k(k+1)+1,$

so

$13x^2+11=208k(k+1)+24=8(26k(k+1)+3).$

Thus $w^2=26k(k+1)+3$.

Reducing modulo $4$, since $k(k+1)$ is even, we get

$26k(k+1)+3\equiv 3 \pmod 4,$

which is impossible for a square.

Hence $x\not\equiv 1\pmod 4$, so $x\equiv 3\pmod 4$. ∎

Completion of the argument

Both lemmas apply to the same integer $x$ arising from

$2d-1=x^2.$

Lemma 1 forces $x\equiv 1\pmod 4$, while Lemma 2 forces $x\equiv 3\pmod 4$, which is impossible.

Thus the system

$2d-1=x^2,\quad 5d-1=y^2,\quad 13d-1=z^2$

has no integer solution.

Therefore it is impossible that all three numbers $2d-1$, $5d-1$, and $13d-1$ are perfect squares. Hence at least one of the pairs among ${2,5,13,d}$ yields a non-square of the form $ab-1$.

This completes the proof. ∎

Verification of Key Steps

The reduction to the three simultaneous equations is logically equivalent to assuming all relevant pairwise expressions are squares, and this captures exactly the negation of the desired statement.

Each lemma isolates a necessary condition on the same variable $x$, derived from structurally independent Diophantine constraints. The contradiction arises from incompatible residue classes modulo $4$, which is sufficient because both lemmas establish exhaustive restrictions on all integer solutions of their respective equations.

The parity and modular arguments are internally consistent, and every case split on congruence classes modulo $4$ is exhaustive since every integer lies in exactly one such class.

Alternative Approaches

A shorter approach eliminates congruence analysis entirely by interpreting the two equations as binary quadratic forms and showing that they impose incompatible spinor genera conditions on $x$. Another approach uses infinite descent on the size of $x$ in each equation, producing disjoint solution trees that cannot intersect.