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Only Finite Integer Solutions to These Quartics

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 \((x,y)\), there does not exist an integer solution to \[x^4+x+1=y^2\]

Problem 2: Prove that for all positive ordered pairs of integers \((x,y)\), there does not exist an integer solution to \[x^4+x+2=y^2\] other than the pair \((1,2)\)

Problem 3: Prove that for all positive ordered pairs of integers \((x,y)\), there does not exist an integer solution to \[x^4+x+7=y^2\] other than the pairs \((1,3)\) and \((2,5)\)


Post your solutions below. Have fun!

Note by Daniel Liu
3 years, 1 month ago

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Problem 1: Note that \(x\) is given to be positive. Hence, the given equation implies that \(y^2 > x^4\), i.e. \( y> x^2 \), i.e. \(y \geq (x^2+1)\). Hence \(y^2\geq x^4 + 2x^2+1\). The given condition then implies \(x^4 + x +1 \geq x^4 + 2x^2 +1\), i.e. \(x \geq 2x^2\). which does not have any positive integral solution. Other two problems can also be nailed down by similar arguments. Abhishek Sinha · 3 years, 1 month ago

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@Abhishek Sinha Yep, that's how I solved it. Daniel Liu · 3 years, 1 month ago

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@Abhishek Sinha Awesome!!I also used similar arguments... Eddie The Head · 3 years, 1 month ago

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[Whispers with ghostly voice] ...bound between squares... Luis Rivera · 3 years, 1 month ago

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@Luis Rivera Correct! Daniel Liu · 3 years, 1 month ago

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I would like to share a few more problems which use similar ideas.

Sreejato Bhattacharya · 3 years, 1 month ago

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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..................... Saurav Pal · 3 years, 1 month ago

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