Posts Tagged ‘quartics’

A Fermat’s Last Theorem near-miss

By Fermat’s Last Theorem, the quartic equation,

x^4+y^4 = z^4

has no non-trivial rational solutions.  In fact, the same can be said for the less strict,

x^4+y^4 = z^2

So how do we explain the near-equalities,

24576^4+48767^4 \approx (49535.000000000006\dots)^4

419904^4 + 1257767^4 \approx (126155.000000000000001\dots)^4

A search for others with z < 8,000,000 will not yield better approximations.  Noam Elkies showed that an identity is behind it, namely,

(192v^8-24v^4-1)^4+ (192v^7)^4= (192v^8+24v^4-1)^4+12(2v)^4

Since the second term on the RHS is small compared to the others, this gives an excellent near-miss to Fermat’s Last Theorem.  Inspired by Elkies’ quartic identity, Seiji Tomita found similar ones as,

(48v^8-12v^4-1)^4 + 2(48v^7)^4 = (48v^8+12v^4-1)^4 + 6(2v)^4

(12v^8-6v^4-1)^4 + 4(12v^7)^4 = (12v^8+6v^4-1)^4 + 3(2v)^4

It can be shown all three belong to the same family.  For arbitrary m let,

b = \frac{8}{m},\;\; d = \frac{3m}{2}

then,

(3m^2 v^8-3m v^4-1)^4+b(3m^2 v^7)^4=(3m^2v^8+3mv^4-1)^4+d(2v)^4

Note that,

bd = 12

In general, these are diagonal quartic surfaces of form,

ax^4+by^4 = cz^4+dt^4

The case {a, b, c, d} = {1, 2, 1, 4}, or the Swinnerton-Dyer quartic surface, was discussed in the previous post and, in the link above, Elsenhans gives first-known but large solutions to other positive {a, b, c, d}.  From Elkies’ and Tomita’s results, it is tempting to speculate that if,

x^4+2y^4 = z^4+4t^4

has a non-trivial solution, then does

x^4+y^4 = z^4+8t^4

have one as well?  Unfortunately, this particular form does not seem to be in Elsenhans’ list.  (For positive {a, b, c, d}, the only other case I know of that has an infinite number of solutions is a = b = c = d = 1.)

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The Swinnerton-Dyer quartic surface

Does the Diophantine equation,

\text{(1)}\,\,\, a^4+2b^4=c^4+4d^4

first suggested by Swinnerton-Dyer, only have a finite number of primitive solutions? Note that, by Sophie Germain’s identity, the RHS factors as,

c^4+4d^4 = \big((c+d)^2+d^2\big)\big((c-d)^2+d^2\big) = (c^2+2cd+d^2)(c^2-2cd+d^2)

Before discussing (1), we can compare it to other quartic Diophantine equations.  Any primitive integral solution to,

\text{(2)}\,\,\, a^4+b^4+c^4+d^4 = e^4

must obey the congruential constraint that there is an addend such that either {d^2+e^2,\, d^2-e^2} is divisible by 5^4.  The smallest was found by Norrie in 1911 as,

(a,b,c,d,e) = (30, 120, 315, 272, 352)

and indeed,

\begin{aligned}d+e &= 272+353 = 5^4\\ -d+e &= -272+353 = 3^4 \end{aligned}

On the other hand, for,

\text{(3)}\,\,\, a^4+b^4+c^4 = d^4

Morgan Ward proved the added constraint that either {c+d, -c+d} is divisible by 2^{10}. The smallest was found by Roger Frye in 1988,

(a,b,c,d) = (95800, 414560, 217519, 422481)

and we can see that,

c+d = 217519+422481 = 2^{10}\, 5^4

Going back to the Swinnerton-Dyer equation, the only solution less than a hundred million (10^7) is,

(a,b,c,d) = (1484801, 1203120, 1169407, 1157520)

found by A. Elsenhans and J. Jahnel in 2004.  On a hunch, I decided to test the values and found that,

a+c = 1484801+1169407 = 2^{15}\,3^4

I don’t know if this is a general congruence valid to all primitive solutions of (1), and an email to the authors revealed neither did they.  In fact, to quote their paper, “... it is unknown whether this equation admits finitely or infinitely many primitive solutions. If their number were actually finite then this would settle a famous open problem in the arithmetic of K3 surfaces…

So many questions, so little time.

Quartic equations

Composite degrees n are better solved by polynomial decomposition.  However, for n = 2^m, there is an alternative method.  Thus, we’ll give two methods to solve the quartic.  Given the depressed quartic,

x^4+px^2+qx+r = 0

Its resolvent cubic is simply,

y^3+2py^2+(p^2-4r)y-q^2=0

Method 1:

Given any root y of the cubic, then,

\begin{aligned} x_{1,2} &=\tfrac{1}{2} \left(\sqrt{y} \pm \sqrt{-(2p+y)-\tfrac{2q}{\sqrt{y}}}\,\right)\\ x_{3,4} &= \tfrac{1}{2} \left(-\sqrt{y} \pm \sqrt{-(2p+y)+\tfrac{2q}{\sqrt{y}}}\, \right)\end{aligned}

For convenience, one can choose the real root y.  Notice that {x_3, x_4} are the negative case of \pm \sqrt{y}.  The formula can easily be proven by squaring the variable y, forming the quadratic in {x_1,\,x_2}, and eliminating y between them using resultants,

\text{Factor}[\text{Resultant}[y^6+2py^4+(p^2-4r)y^2-q^2,\,x^2+xy+\frac{y^3+py-q}{2y},\,y]]\\ = (x^4+px^2+qx+r)^3

proving any root of the sextic (a cubic in y^2) will suffice.

Method 2:

Given the three roots {y_1, y_2, y_3} of the resolvent cubic, then,

\begin{aligned} x_{1,2} &= \frac{\sqrt{y_1} \pm (\sqrt{y_2}+\sqrt{y_3}\,)}{2}\\ x_{3,4} &= \frac{-\sqrt{y_1} \pm (\sqrt{y_2}-\sqrt{y_3}\,)}{2}\end{aligned}

where signs are chosen such that,

[1]  \sqrt{y_1}\sqrt{y_2}\sqrt{y_3} = -q

Proof:  Expanding,

(x-x_1)(x-x_2)(x-x_3)(x-x_4) = 0

using the formulas, we get,

x^4-\frac{1}{2}(y_1+y_2+y_3)x^2-\sqrt{y_1}\sqrt{y_2}\sqrt{y_3}\,x\\+\frac{1}{16}(y_1^2+y_2^2+y_3^2-2(y_1y_2+y_1y_3+y_2y_3)) = 0

From [1] and the properties of elementary symmetric polynomials, we find that it is in fact equal to,

x^4+px^2+qx+r = 0

A similar method was used to solve the octic x^8-x^7+29x^2+29 = 0 in this post, though now one had to take square roots of its resolvent septic.

Cubic equations

Recall that solving the quadratic equation,

x^2+ax+b = 0

gives,

x = \frac{1}{2}\,(-a \pm \sqrt{y}\,)

where y is the linear,

y = a^2-4b

The solution to the cubic equation is simply a generalization. Given,

x^3+ax^2+bx+c=0

then,

x =\frac{1}{3}\,(-a+y_1^{1/3}+y_2^{1/3})

where the y_i are now the two roots of the quadratic,

y^2+(2a^3-9ab+27c)y+(a^2-3b)^3 = 0

Notice that the constant term of its resolvent is a perfect cube.  For prime degrees, if the irreducible equation is solvable, then it can be solved analogously. For depressed quintics (an easy linear transformation to make a = 0), then,

x = \frac{1}{5}\,(y_1^{1/5}+y_2^{1/5}+y_3^{1/5}+y_4^{1/5})

where the y_i are now the four roots of its resolvent quartic. For non-binomial quintics, this quartic will have a constant term that is a perfect fifth power. And so on for other prime degrees.

Shanks’ approximation to pi

In “Pi Approximations“, line 58, Weisstein mentions one by Shanks (1982) that differs by a mere 10^{-82} as,

\pi \approx \frac{6}{\sqrt{3502}}\, \ln(2u)

where u is “…a product of four simple quartic units”. Frustratingly, he doesn’t give u but I eventually found the primary source online. Hence,

u = \big(a+\sqrt{a^2-1}\big)^2 \big(b+\sqrt{b^2-1}\big)^2 (c+\sqrt{c^2-1}) (d+\sqrt{d^2-1})

where,

\begin{aligned}a &= \tfrac{1}{2}\, (23+4\sqrt{34}\,)\\ b &= \tfrac{1}{2}\, (19\sqrt{2}+7\sqrt{17}\,)\\ c &= 429+304\sqrt{2}\\ d &= \tfrac{1}{2}\,(627+442\sqrt{2}\,) \end{aligned}

with a slight modification by this author since Shanks didn’t realize the first two quartic factors were in fact squares.  (The product of the last two factors is also a square.)

A cute thing about these numbers is that their defining polynomials are palindromic, the same read forward or backward.  For example, the first factor (unsquared) is the root of,

x^4-46x^3-13x^2-46x+1 = 0

Author’s note:  Daniel Shanks (1917-96) was a mathematician best known as the first to calculate pi up to 100,000 decimal places, as well as for his book, Solved and Unsolved Problems in Number Theory.

In general, Shanks’ approximation belongs to the family,

\begin{aligned} e^{\pi\sqrt{2m}} &\approx \left(\tfrac{\eta(\frac{1}{2}\sqrt{-2m})}{\eta(\sqrt{-2m})}\right)^{24}\\ &\approx 2^6 x^{k}\end{aligned}

where \eta is the Dedekind eta function and, for m a positive odd integer, then x is an algebraic integer that is the root of an equation P(x) with palindromic (if unsigned) coefficients.  For appropriate k, then P(x) has degree equal to the class number h(-2m).  Furthermore, it is solvable in radicals.  

For example, given prime m, with 2m = {10, 14, 26} which has class number 2, 4, 6, respectively, then k = 12 and,

\begin{aligned} x_{10}&\; \text{is a root of}\, x^2-x-1 =0\\ x_{14}&\;\text{is a root of}\, x^4-2x^3+x^2-2x+1 = 0\\ x_{26}&\;\text{is a root of}\, x^6-2x^5-2x^4+2x^2-2x-1 = 0 \end{aligned}

and so on.  It then is a simple matter to take the natural logarithm of both sides to bring down pi and have a relation of form,

\pi \approx \frac{1}{\sqrt{2m}} \ln(2^6 x^{k})

Shanks chose 2m = 3502 since d = 4(2m) is the largest fundamental discriminant d divisible by 4 with class number h(-d) = 16.  Here is a list of of d with small class number.  You can calculate it (among many other things) in www.wolframalpha.com simply as,

ClassNumber[Sqrt[-d]]