Some guidance questions before you tackle Diophantine Quartic. If you can manage to prove these, you will be able to solve my question with no problem.
Problem 1: Prove that for all positive ordered pairs of integers , there does not exist an integer solution to
Problem 2: Prove that for all positive ordered pairs of integers , there does not exist an integer solution to other than the pair
Problem 3: Prove that for all positive ordered pairs of integers , there does not exist an integer solution to other than the pairs and
Post your solutions below. Have fun!
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Problem 1: Note that x is given to be positive. Hence, the given equation implies that y2>x4, i.e. y>x2, i.e. y≥(x2+1). Hence y2≥x4+2x2+1. The given condition then implies x4+x+1≥x4+2x2+1, i.e. x≥2x2. which does not have any positive integral solution. Other two problems can also be nailed down by similar arguments.
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Awesome!!I also used similar arguments...
Yep, that's how I solved it.
[Whispers with ghostly voice] ...bound between squares...
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Correct!
I would like to share a few more problems which use similar ideas.
RMO 2000 Problem 2
RMO 2012 Region 1 Problem 6
We are given that y^2 > x^4+x+1.............. y^2 > x^4.............. y > x^2................. It is also given that x is a positive integer............. So, y>=x^2+1.............
y^2>=x^4+2x^2+1......................
So, y can not be equal to x^4+x+1.....................