Root Flipping, Vieta Jumping, and Low-Degree Curves

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Explanations 5

Vieta Jumping By now, Vieta jumping has become a standard technique for solving some particular type of olympiad number theory problems. It is also known as Root Jumping or Root…
By amaterasu, Catherine Yeo
(IMO 200720072007, Problem 555). Let aaa and bbb be positive integers so that 4ab−14ab − 14ab−1 divides (4a2−1)2(4a^2 − 1)^2(4a2−1)2. Prove that a=ba = ba=b. In short: If 4ab−14…
By amaterasu, Jiazhen Tan
Prove that, for any positive integer kkk, there are positive integers a,b>1a, b > 1a,b>1 such that k=a2+b2−1abk=\dfrac{a^2+b^2-1}{ab}k=aba2+b2−1
By amaterasu
Let a,ba,ba,b be positive integers such that ab−1ab-1ab−1 divides a2+b2a^2+b^2a2+b2. Prove that their quotient is 555. Being the lazy person I am, I'll just outline my attempt (…
By amaterasu, Jiazhen Tan
Let a,b,a, b,a,b, and ccc be positive integers such that 0<a2+b2−abc≤c0 < a^2+b^2-abc \leq c0<a2+b2−abc≤c. Prove that a2+b2−abca^2+b^2-abca2+b2−abc is a perfect square. WLOG ass…
By amaterasu, Catherine Yeo

Problems 1

[Romania TST 200420042004] Find all integer values the expression a2+b2+1ab−1\frac{a^2+b^2+1}{ab-1}ab−1a2+b2+1 can assume for ab≠1ab\neq 1ab=1 where a, b are natural numbers.