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!

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TopNewestProblem 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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– Daniel Liu · 3 years, 1 month ago

Yep, that's how I solved it.Log in to reply

– Eddie The Head · 3 years, 1 month ago

Awesome!!I also used similar arguments...Log in to reply

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

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– Daniel Liu · 3 years, 1 month ago

Correct!Log in to reply

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

RMO 2000 Problem 2

RMO 2012 Region 1 Problem 6

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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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