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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, 8 months 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, 8 months ago

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Yep, that's how I solved it.

Daniel Liu - 3 years, 8 months ago

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Awesome!!I also used similar arguments...

Eddie The Head - 3 years, 8 months ago

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[Whispers with ghostly voice] ...bound between squares...

Luis Rivera - 3 years, 8 months ago

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Correct!

Daniel Liu - 3 years, 8 months ago

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

Sreejato Bhattacharya - 3 years, 8 months 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, 8 months ago

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