Today I was reading J.H.E. Cohn's Eight diophantine equations (1966). On p. 158 he comes across the equation $y^2 = a^3 + 3a$ for odd values of $a$ and writes that this is equivalent to $x^3 + (x+1)^3 = (\frac y2)^2$ for $x = \frac{a-1}2$. He then claims that it is known that this equation has the only integral solutions $x = 0$ and $x = 1$ and refers to Dickson's History, vol. II, p. 580. Dickson refers to Moret-Blanc (1882), who reduces the claim to the diophantine equation $3u^4 - v^4 = 2$ with the obvious solution $u = 1$. Then he states that Lucas has shown that this is the only solution in integers; Lucas does give infinitely many solutions, one of which is $z = 1$, but he does not solve the equation completely.
sage can compute the integral points on $y^3 = x^3 + 3x$ in a couple of seconds (there are four, namely $x =0$, $x = 1$, $x = 3$ and $x = 12$), but now I'm wondering a) whether this was known when Cohn wrote his paper in 1966, and b) whether there is a sufficiently nice solution of $x^3 + (x+1)^3 = z^2$ that can be done by hand. The tricks in my bag don't seem to work.
P.S. I now found that in an addendum to his article, Cohn mentions the gap and writes that a full proof of the claim was given by Ljunggren in his thesis, available to readers with Norwegian IPs here.
