Prime-support bound in two-isogeny descent ($r\le \nu (b)+\nu (a^2-4b)$)

For integral $a,b$ with $b(a^2-4b)\ne 0$, every square class in the image of the two-isogeny descent map on $y^2=x(x^2+ax+b)$ has a signed square-free representative supported on primes dividing $b$. At a prime not dividing $b$, a positive valuation of $x$ must be even from the curve equation; a negative valuation is even because the cubic leading term determines $2v(y)=3v(x)$. Thus its image has at most $2^{\nu (b)+1}$ elements. Apply this also to the two-isogenous curve and use the square-class index formula for two-isogeny descent to obtain $r\le \nu (b)+\nu (a^2-4b)$, where $\nu$ counts distinct prime divisors of the nonzero absolute value.