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Can someone check pls if,

$$852 + 3017 x - 1104 x^2 + 2009 x^3 - 3362 x^4=y^2$$

has a rational point? (This arose in an equal sums of like powers problem.)

P.S. I've checked $x=p/q$ for $\text{Abs}\,(p,q)<200$, but didn't find anything. :(

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Have you tried SAGE mathoverflow.net/questions/42016/…? –  Amzoti Sep 27 '13 at 3:04
    
If it needs some C++ coding... :( –  Tito Piezas III Sep 27 '13 at 3:31
    
No, it needs no coding, just enter the equation. They show examples. Is it a free Computer Algebra System. –  Amzoti Sep 27 '13 at 3:36
    
The polynomial in $\;x\;$ on the left is one degree too high for that equation to represent an elliptic curve –  DonAntonio Sep 27 '13 at 4:53
    
If a 4th deg polynomial to be made a square has a non-trivial rational point, it is quite easy to use a birational transformation to transform it to the 3rd deg Weierstrass form, hence an elliptic curve. (The problem is to find that initial rational point, if it has any.) –  Tito Piezas III Sep 27 '13 at 4:57

1 Answer 1

up vote 1 down vote accepted

There may be faster ways, but this is what I have found.

Assume $x=p/q$ with $p,q$ coprime integers and $q$ positive. Multiply the curve equation by $q^4$ to make its left-hand side an integer: $$F(p,q)=-3362 p^4 + 2009 p^3 q -1104 p^2 q^2 + 3017 p q^3 + 852 q^4=(q^2 y)^2$$ Therefore $q^2 y$ must be an integer to solve the equation, and the right-hand side must be a square in $\mathbb{Z}$.

Assume $5\mid q$, then $F(p,q)\equiv-2 p^4\pmod{5}$, which is not a square unless $5\mid p$, but the latter would contradict the lowest-terms requirement. So $5\not\mid q$, hence any solution's $x$ can be represented in $\mathbb{Z}_5$ (the $5$-adic integers), and $F(x,1)$ must be a square in $\mathbb{Z}_5$.

Using Pari/GP, I verified that $F(x,1)$ is never a square modulo $5^4$, so the original equation has no rational solutions.

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Thanks. Last one, pls. Can you check with ratpoints if -3166834256 + 900727966 x - 162228207 x^2 + 11996914 x^3 + 53215 x^4 = y^2 has a rational solution of small height?. If so, these will yield new solutions to $a^4+b^4+c^4 = 1$ using elliptic curves distinct from the ones already found. –  Tito Piezas III Oct 19 '13 at 22:31
    
@TitoPiezasIII: r.ratpoints([-3166834256,900727966,-162228207,11996914,53215],H) immediately returns an empty list even for very large $H$, so there must be an easy proof that no rational $x,y$ fulfill the equation. Indeed $11\not\mid q$ and $F(x,1)$ is never a square $\bmod\ 11$, along the lines described. –  ccorn Oct 19 '13 at 22:54
    
Rats... Thanks though. –  Tito Piezas III Oct 19 '13 at 23:14
    
If you are interested, only six $|m,n|<200$ are known so far (Bremner recently found a sixth one) that yield $a^4+b^4+c^4 = 1$. This post should explain more. –  Tito Piezas III Oct 19 '13 at 23:19

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