2
$\begingroup$

In 1643, Fermat asked Frenicle et al to find a special Pythagorean triple $a,b,c$ such that for $n=1$,

$$a+nb = r_1^2\\ a^2+b^2 = r_2^4\tag1$$

Equivalently,

$$\color{blue}{\big((p^2-q^2)^2-(2pq)^2\big)}+n\color{brown}{\big(2(p^2-q^2)(2pq)\big)} = r_1^2\tag2$$

This has polynomial solution $p,q = n^2+2, 2n.$ (And an infinite more, since it can be birationally transformed to an elliptic curve.) But this simple solution yields negative $a$ for $n=1,2,3,4$.

Fermat stated that the smallest positive answer for $n=1$ was,

$$n,a,b = 1,\;4565486027761,\; 1061652293520$$

Q: What is the smallest positive solution for $n=2,3,4$?

For $n=2$, I think it is,

$$n,a,b = 2,\;10386304269597791463121,\; 3672193546323671330640$$

though I am not sure.

$\endgroup$
1
$\begingroup$

This is a really nice question! I think your answer for $n=2$ is correct.

Equation $(2)$ can be shown to be equivalent to the elliptic curve \begin{equation*} v^2=u^3-(n^2+1)u \end{equation*} with \begin{equation*} \frac{p}{q}=\frac{nu-v}{n^2+1-u} \end{equation*}

The elliptic curve has one finite torsion point at $(0,0)$, though I have not attempted to prove this. There is, also, a point when $u=-1$ and $v=n$, so that the curve always has rank greater than zero.

I wrote a Pari-gp program which uses Denis Simon's ellrank code to determine the rank and generators of the curve. It then searched through rational points to find a minimum positive solution. The results are

n = 1

Rank of curve = 1

Minimum = [4565486027761, 1061652293520]

n = 2

Rank of curve = 1

Minimum = [10386304269597791463121, 3672193546323671330640]

n = 3

Rank of curve = 1

Minimum = [166911677107794033180761521, 383496393351054937416817200]

n = 4

Rank of curve = 2

Minimum = [230903401, 2224677000]

n = 5

Rank of curve = 1

Minimum = [104041, 679320]

n = 6

Rank of curve = 1

Minimum = [53641, 148200]

n = 7

Rank of curve = 1

Minimum = [3744841, 6868680]

n = 8

Rank of curve = 2

Minimum = [876353497971071281, 193103317517647440]

n = 9

Rank of curve = 3

Minimum = [152401, 142800]

$\endgroup$
  • $\begingroup$ Thanks! Good to know I got $n=2$ right, and I'm glad you extended it to $n>4$. I checked with my parameterization of $(2)$ above and, after removing common factors, our solutions for $n=5,6,7$ are the same. However, for $n=8$, the parametrization yields the smaller, $$a,b = 771841,\,1082400$$ while yours is, $$a,b = 876353497971071281,\, 193103317517647440$$ What could be the reason? $\endgroup$ – Tito Piezas III Mar 1 '16 at 15:48
  • $\begingroup$ I did a small Sage script following Allan's method and was able to find $a, b = 771841, 1082400$ by the same method. It correspond to the point $(-1,8)$ on the elliptic curve which is among the generators returned by Sage. It could be that Pari and Sage gives different sets of generators? $\endgroup$ – Jesper Petersen Mar 1 '16 at 18:50
  • $\begingroup$ The problem was that I used the wrong variable name in one line of code, but this code gave your solutions for $n=1$ and $n=2$ so I assumed it was correct. Above solutions are unchanged apart from $n=4$ gives $45241, 26520$, and $n=8$ actually gives $166873, 272136$ as smallest solution. $\endgroup$ – Allan MacLeod Mar 2 '16 at 7:28
  • $\begingroup$ I curious as to what points on the elliptic curves correspond to the solutions for $n = 2,3$. I am able to reproduce the solutions except for these two cases. $\endgroup$ – Jesper Petersen Mar 2 '16 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.