Given this diophantine equation: $$16r^4+112r^3+200r^2-112r+16=s^2$$
Wolfram alpha says the only solutions are $(r,s)=(0,\pm4)$
How would one prove these are the only solutions? Thanks.
Asked
Active
Viewed 296 times
2
-
1Hint: $P(r)=s^2$ divides by $8,$ but not by $16.$ – Lucian Dec 06 '15 at 09:09
1 Answers
3
They are NOT the only integer solutions.
We also have $(-2, \pm 20)$ and $(-8, \pm 148)$.
Basically, the quartic is birationally equivalent to the elliptic curve \begin{equation*} V^2=U^3-142U^2+5040U \end{equation*} with \begin{equation*} r=\frac{7U-V-504}{4(63-U)} \end{equation*}
The elliptic curve has $3$ finite torsion points of order $2$ at $(0,0)$, $(70,0)$ and $(72,0)$, which give $r=0,1/2,-2$. The curve has rank $1$ with a generator $(56,112)$, which gives $r=-8$.
Rank $1$ implies an infinite number of rational points which lead to an infinite number of rational values of $r$. I have not checked whether any more of these are integers.
Allan MacLeod
- 1,709
- 1
- 10
- 6
-
1is there a good source for this method of solving diophantine equations with elliptic curves? – AnotherPerson Dec 27 '15 at 03:59