Square numbers within an arithmetic progression

[apple2357]

Been playing with a spreadsheet on excel and exploring under what conditions an arithmetic progression will generate square numbers.



Just working within natural numbers starting with a=1 and using d=1,2,3 etc - we appear to stumble upon square numbers and i think they will continue to be generated. In fact, once i started to mess around i discovered the case a=8, d=3 ( checked up to 150) didn't generate a square number.
I have a conjecture that if a is a square number then the AP will contain further square numbers.
However, if a is not square, you can still generate square numbers ( a=5, d=2) but not guaranteed to.

So, how do i know which APs defined will generate square numbers and which will not? My hunch tells me if a is square then you will definitely get square numbers. Any ideas on a proof?

So far i have thought about (a+d)^2 and under what condition this might generate an AP term but not sure where to go next!

 

[JeffM]

You know, I am sure, that this kind of conjecture can be amazingly hard to prove or disprove.

In what follows j, k, and m are all positive integers and

[MATH]n = j + km[/MATH].

As a start, I observe from your spread sheet

[MATH]j = 1,\ k = 3, \text { and } m = 1 \implies n = 4 \implies \sqrt{n} = 2 \in \mathbb Z.[/MATH]
[MATH]j = 1,\ k = 4, \text { and } m = 2 \implies n = 9 \implies \sqrt{n} = 3 \in \mathbb Z.[/MATH]
[MATH]j = 1, k = m - 2, \text { and } m \ge 3 \implies n = 1 + (m - 2)m = m^2 - 2m + 1 = (m - 1)^2 \implies \sqrt{n} \in \mathbb Z.[/MATH]
This just formalizes what your spread sheet screamingly suggests and generalizes it to all cases where j = 1. It could be a base case for a proof by induction.

I am not sure what to do next.

 

[Cubist]

[MATH]j = 1, k = m - 2, \text { and } m \ge 3 \implies n = 1 + (m - 2)m = m^2 - 2m + 1 = (m - 1)^2 \implies \sqrt{n} \in \mathbb Z.[/MATH]
This just formalizes what your spread sheet screamingly suggests and generalizes it to all cases where j = 1. It could be a base case for a proof by induction.

I think this proves the case where j is a perfect square (this is a very small extension to @JeffM's post)...

[MATH]j =x^2: x \in \mathbb Z, k = m - 2x, \text { and } m \gt 2x \implies n = x^2 + (m - 2x)m = m^2 - 2mx + x^2 = (m - x)^2 \implies \sqrt{n} \in \mathbb Z.[/MATH]

 

[JeffM]

I ran the j = 8, m = 3 case up to k = 5000 without finding a square resultant. Of course, that does not prove that there are AP sequences that contain no perfect squares, but it does suggest that the conjecture is not trivial to disprove by numerical experimentation.

 

[JeffM]

[MATH]j = 8 \text { and } m = 3.[/MATH]
[MATH]\text {ASSUME } \exists \text { integer } k \text { such that } \sqrt{8 + 3k} \in \mathbb Z.[/MATH]
[MATH]\text {Let } p \text { and } q \text { be any positive integers such that } p = q + \sqrt{8 + 3k}.[/MATH]
[MATH]\therefore p^2 - 2pq + q^2 = 8 + 3k \implies k = \dfrac{p^2 - 2pq + q^2 - 8}{3} = \dfrac{p^2 - 2pq + q^2 - 6 - 2}{3}.[/MATH]
[MATH]\therefore \exists \text { positive integers } \alpha \text { and } \beta \text { and non-negative integers } \gamma \text { and } \delta[/MATH]
[MATH]\text {such that } \gamma \le 2 \ge \beta \text { and } p = 3 \alpha + \gamma \text { and } q = 3 \beta + \delta.[/MATH]
[MATH]\therefore p^2 = 9 \alpha^2 + 6 \alpha \gamma + \gamma^2 \text { and } q^2 = 9 \beta^2 + 6 \beta \delta + \delta^2 \text { and } - 2pq = - 18 \alpha \beta - 6 \alpha \delta - 6 \beta \gamma - 2 \gamma \delta.[/MATH]
[MATH]\therefore k = \dfrac{p^2 - 2pq + q^2 - 6 - 2}{3} = \dfrac{3 * \text { some integer} + \gamma^2 + \delta^2 - 2 \gamma \delta - 2}{3} = \text {some integer} + \dfrac{(\gamma - \delta)^2 - 2}{3}.[/MATH]
[MATH]\gamma - \delta = 0, \ \pm 1, \text { or } \pm 2 \implies (\gamma - \delta)^2 = 0, \ 1 , \text { or } 4 \implies [/MATH]
[MATH](\gamma - \delta)^2 - 2 = - 2, \ - 1, \text { or } 1 \implies \dfrac{(\gamma - \delta)^2 - 2}{3} \not \in \mathbb Z \implies[/MATH]
[MATH]k \text { is not an integer, which contradicts the assumption.}[/MATH]
[MATH]\text {There is no integer } k \text { such that } \sqrt{8 + 3k} \in \mathbb Z.[/MATH]
We have proved that the sequence defined by 8 + 3k contains no perfect squares.

So we know that some sequences in arithmetic progression contain perfect squares, and some do not.

What are the distinguishing characteristics between the two types of arithmetic progression?

 

[apple2357]

Just been thinking about this a little more... Its clear all square numbers either leave 0 or 1 remainder after being divided by 4.
( congruent to 0 or 1, modulo 4). This i feel must put restrictions on the type of AP that will not generate square numbers?

 

[JeffM]

Just been thinking about this a little more... Its clear all square numbers either leave 0 or 1 remainder after being divided by 4.
( congruent to 0 or 1, modulo 4). This i feel must put restrictions on the type of AP that will not generate square numbers?

@apple2357

I suspect you are on a promising path. I have not followed up, partly because I have had other demands on my time and partly because my idle thoughts on this problem have been along the lines of generalizing what I previously did without much result.

[MATH]\sqrt{j + km} \in \mathbb Z \implies \exists \text { integer } p \text { such that } p = m + \sqrt{j + km} \implies [/MATH]
[MATH]p^2 - 2pm + m^2 = j + km \implies k = \dfrac{p^2 - 2pm + m^2 - j}{m} = m - 2p + \dfrac{p^2 - j}{m}.[/MATH]
If that fraction is not reducible to an integer, then k is not an integer, which gives the contradiction.