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Statement

Theorem 3 (p. 49). Let $2=p_1,p_2,\ldots$ be the primes in order and let $\varepsilon>0$. There is a number $s_0(\varepsilon)$, effectively computable in terms of $\varepsilon$, such that for every integer $s>s_0(\varepsilon)$ there are positive integers $k_1$ and $k_2$ with

$$k_1<\exp\bigl(2(s\log s)^{1/2}\bigr),\qquad k_2<\exp\bigl((s\log s)^{1/2}\bigr),$$

such that

• the equation $x-y=k_1$ has at least $\exp\bigl((4-\varepsilon)(s/\log s)^{1/2}\bigr)$ solutions in positive integers $x,y$ with $P(xy)\le p_s$, and
• the equation $x-y=k_2$ has at least $\exp\bigl((2-\varepsilon)(s/\log s)^{1/2}\bigr)$ solutions in coprime positive integers $x,y$ with $P(xy)\le p_s$.

Here $P(n)$ is the greatest prime factor of $n$. The authors describe Theorem 3 (p. 49) as showing that in Theorem 4 one of $x$ and $y$ can be fixed at the cost of replacing the exponent $4-\varepsilon$ by $2-\varepsilon$.

Proof pointer

Pp. 49--51. Both parts apply Lemma 3 (p. 40) with $l=2$ and $f(x)=(\log x)/2$: with $c=1$ and $N=\lfloor\exp((2-\delta)(s\log s)^{1/2})\rfloor$ for the first part, and with $c=4$ and $N=\lfloor\exp((1-\delta)(s\log s)^{1/2})\rfloor$ for the second. The pairs $a_i$, $a_i+b$ it supplies are taken as $y$ and $x$ with $k=b$, and the prime number theorem gives $P(xy)\le p_s$. For the coprime part, each solution is divided by $\gcd(x,y)$, which divides $k$; the divisor bound of Hardy and Wright (Theorem 317) limits the number of resulting differences $k/d$, so one of them keeps enough solutions.

Read depth

Claims checked: the statement was read clause by clause on the page image of the print. The proof was followed for the outline above and is not independently verified.

Dependencies

• Lemma 1 (p. 39), through Lemma 3 (p. 40).

Source. P. Erdős, C. L. Stewart and R. Tijdeman, Some diophantine equations with many solutions, Compositio Mathematica 66 (1988), 37--56; the edition read is named on the source card.

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