From the Wikipedia page about Vieta jumping:

Emanouil Atanassov, Bulgaria, solved the problem [assumed to be the most difficult one on the 1988 International Mathematics Olympiad] in a paragraph and received a special prize.

The reference links to this page. It does not mention anything about that paragraph. Does anyone have a copy of the actual paragraph written by him to solve the problem?

  • 8
    $\begingroup$ Perhaps dynamicmathematicslearning.com/appendix-MKSiu-IMO2016.pdf $\endgroup$
    – sand1
    Aug 11, 2020 at 13:31
  • 2
    $\begingroup$ There's a video by Numberphile starring Zvezdelina Stankova (the other Bulgarian on the Olympiad that year) called "The Notorious Question Six". She basically shows his solution (using induction and quadratic polynomial/Vieta's formulas). It's really quite elegant. $\endgroup$
    – Nenorr
    Aug 7, 2022 at 15:39

1 Answer 1


I just want to point out that Don B. Zagier's one-paragraph solution to this notorious problem is more or less well-known:

Let $m$ be the quotient. We may suppose that $(a, b)$ is a minimal positive solution of the equation $(a^2 + b^2) = m(ab + 1)$
(i.e. one with the smallest value of $a + b$) for this value of $m$. Without loss of generality, suppose that $b ≥ a ≥ 0$, and set $b' = ma - b$. Then $a^2 = bb' + m$. If $b'$ is positive, then this equation implies $b' = (a^2 - m)/b < a^2/b < b$, and $(a, b')$ is a smaller positive solution of the equation of which $(a, b)$ was supposed to be the minimal solution. If $b'$ is negative, then $m = a^2 - bb' ≥ a^2 + b > b > ma ≥ m$, a contradiction. Hence $b' = 0$ and $m = a^2$.

Note: This solution is just the explicit result of applying reduction theory (specifically, Sätze 1 and 2 of Section 13 of my book on quadratic fields) to the quadratic form $x^2 + mxy + y^2$, which is the unique reduced quadratic form in its equivalence class.

(cf. http://www-groups.mcs.st-andrews.ac.uk/~john/Zagier/Solution1.3.html)

Maybe E. Atanassov's solution was similar to the above solution by Zagier.


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