# Squares in an Arithmetic Progression

Let $P(x;a,b) := \{an+b, 0\leq n \leq x \}$ denote an arithmetic progression. Further let $A(x;a,b)$ denote the number of elements of $P(x;a,b)$ that are squares. It's an old conjecture of Rudin that $A(x;a,b) \ll x^{1/2}$. Less ambitiously, Erdős posed the problem of showing that $A(x;a,b) = o(x)$. This was proven by Szemeredi around 1974 (amusingly the paper is only a few sentences).

Here is Szemeredi's proof: If the theorem was false then we could find arbitrarily large arithmetic progressions composed of at least $\delta>0$ percent squares. Then invoking the 4 case of Szemeredi's (most well-known) theorem we have that there must be a length 4 arithmetic progression consisting of only squares. However, this contradicts an old theorem of Euler.

While this proof is slick, it is natural to want to avoid having to use anything as powerful as Szemeredi's theorem. I recently I ran across the paper "On the Number of Squares in an Arithmetic Progression" by Saburo Uchiyama (Proc. Japan Acad. 52, no. 8 (1976), 431-433). The complete paper is freely available here. The paper claims to give a simple and self-contained solution to Erdős' question (that $A(x;a,b) = o(x)$). In fact, the proof given is so short that I will repost it in its entirety:

Now I don't follow the claim that "This clearly proves (1)." (where (1) is the claim that $A(x;a,b) = o(x)$). Certainly when $a=o(x)$ this gives the desired result, but why does this work when $a$ is large?

Since this is so short and simple (and I have never seen the argument cited anywhere) I am skeptical, however perhaps I am missing something obvious.

(Perhaps, it should be pointed out that there is a more recent approach to the problem that yields better quantitative bounds that goes through Falting's theorem due to Bombieri, Granville and Pintz. In fact, their analysis shows that the case when $a$ is large compared to $x$ is the 'hard case', which raises further suspicion of the above argument.)

• Picture link is broken – Felipe Voloch Mar 30 '12 at 20:01
• Isn't $x$ growing but $a$ fixed? On the other hand, the question seems trivial unless you want to the implied constants in $o()$ and $\ll$ to be INDEPENDENT of $a,$ and it certainly seems that the paper does not prove this, for the reason you give. – Igor Rivin Mar 30 '12 at 20:22
• What about a lower bound to $A(x;a,b)$? Is anything known? – kodlu Mar 9 '14 at 23:09
• Obviously, Uchiyama did not understand the true question of Erdös, which is trivial if $a$ is fixed. But if he saw the problem formulated as in this post, he is excusable. Saying that the implied constant is absolute, or something like this, would have help. As it is formulated, we are induced to believe that it may depend on $a,b$: because of your title first, "squares" (plural) in "an arithmetic progression" (singular), as if the arithmetic progression, hence $a$ and $b$, where fixed. – Joël Oct 12 '17 at 15:29
• Also because of your notation as well $A(x;a,b)$ and not $A(x,a,b)$, as if $a$ and $b$ were parameters (fixed) and not variables. – Joël Oct 12 '17 at 15:30

The conjecture of Rudin is about the maximum of $A(x; a,b)$, taken over all values of $a$ and $b$. Not with just keeping $a$ and $b$ fixed and letting $x$ get large. So, like you said, if $a$ is not $o(x)$, the proof of Uchida doesn't go through.
• Just to recap: I was aware that both Rudin's and Erdos' problems required uniformity in $a$ and $b$. My question was basically "I don't understand how Uchiyama's argument gives uniformity in $a$." The answer turns out to be: "it doesn't." The statement (in french) of Erdos' problem can be found in bolyai.math-inst.hu/~p_erdos/1963-14.pdf (problem 16). It seems to not be completely clear on the uniformity issue (although its hard to imagine that it never occurred to the author or referee that Erdos was asking for a uniform estimate). – Mark Lewko Mar 31 '12 at 3:05