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I gritted my teeth and dove into some sieve theory. In his 1978 article On the Problem of Jacobsthal in Demonstratio Mathematica, Iwaniec presents two Lemmas to prove his main result (leading to $j(n)$ is $O((r\log r)^2)$), and I am having trouble already with the simpler Lemma. However, what concerns me is what seems to be a false derivation of an inequality. I transcribe the trouble spot and then give the setup. From earlier context, I believe $l(d)=n$; from the article:

"From (2) we have $f(q) \geq g(p)$, where $p = l(q)$. Hence

$$\sum_{d \mid Q} \frac{\delta_{l(d)}}{\prod_{q \mid d} (f(q)-1)} \geq \sum_{d \mid Q} \frac{\delta_{l(d)}}{\prod_{q \mid d} (g(l(q))-1)} = \sum_{n \mid P} \frac{\delta_n}{\prod_{p\mid n}(g(p)-1)}$$

This completes the proof of (4)."

I think if $x \geq y \gt 0$ then $stuff/x \leq stuff/y$ but the above has it reversed. I may be misunderstanding things, but it looks like this breaks the proof.

Here is the setup. The target quantity depends on the square free number $Q$, and estimating the size of some sets with error depending on a positive multiplicative function $f(d)$. Iwaniec is going to shift or transfer the problem over to a specific square free number $P$ and associated positive multiplicative function $g(n)\geq 1$. He posits that $P$ and $Q$ are multiplicatively bijective: there is a bijection $l()$ between the divisors of $Q$ and the divisors of $P$ so that $l(dk)=l(d)l(k)$. He also assumes as (2) that $g(l(d)) \leq f(d)$ for all divisors $d$ of $Q$. I've used $\delta$ instead of the symbol he uses, but I believe they represent positive numbers. I've also used the product symbol $\prod$ for what he uses, but I think it is the right choice. (If the values $\delta$ are all negative, then the derivation is correct, but the conclusion seems too weak, namely the lower bound of a nonnegative quantity is a complicated expression of a negative number, which I do not understand.)

If need be, I can mail a PDF which is 7 small pages long with the excerpt above on page 4. Here are my questions:

Have I fumbled, and the trouble spot is not a trouble spot at all?

If I am misunderstanding something about this Lemma, what is it and can someone explain it to me?

If I have found a problem, can someone fix the Lemma and tell me what the fix is?

If the Lemma can't be fixed, can Iwaniec's result be proven without the Lemma?

It is my hope that some expert eyes will review the pages of the article and assure me that things will turn out alright ( in a fashion I find mathematically convincing ).

Gerhard "That's Enough Questions For Now" Paseman, 2016.07.31

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    $\begingroup$ Hahaha, Iwaniec on the Internet. Good luck with that. $\endgroup$
    – Felipe Voloch
    Commented Aug 1, 2016 at 4:32
  • $\begingroup$ If he responds on MathOverflow, I will have you to thank for that, Felipe. Gerhard "Remember The John Tate Summons?" Paseman, 2016.07.31. $\endgroup$
    – Gerhard Paseman
    Commented Aug 1, 2016 at 4:43
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    $\begingroup$ That was easy. Tate's office was down the hall from mine at the time. I told him about the question and, when he went to MO and answered, it was too good for me to say anything. I can't do that with Iwaniec, sorry. $\endgroup$
    – Felipe Voloch
    Commented Aug 1, 2016 at 4:59
  • $\begingroup$ No problem. Call up someone on MathOverflow at Rutgers. Gerhard "Hopefully They're Already Reading This" Paseman, 2016.07.31. $\endgroup$
    – Gerhard Paseman
    Commented Aug 1, 2016 at 5:01
  • $\begingroup$ As another hint (element of confusion), the Delta's are defined so that for this part of the proof $\delta_m \geq \sum_{n \mid m} \mu(n)$ for all $m \mid P$. I am guessing the $m$ are positive and $\mu()$ is the Möbius function, which would make the Delta's all nonegative, with $\delta_1 \geq 1$. Gerhard "Can Use Some Expert Help" Paseman, 2016.07.31. $\endgroup$
    – Gerhard Paseman
    Commented Aug 1, 2016 at 5:47

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