## 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.