1. Generate at Least 7 Different Arithmetic Progressions
Let the seed be (s), which is a whole number from (0) to (10), and let the constant difference be (d), which is a whole number from (1) to (10). The formula for the (n)-th term is
[
x_n=s+nd
]
where (n=0,1,2,3,\ldots).
Here are seven different arithmetic progressions:
Progression 1: (s=2), (d=3)
[
2,5,8,11,14,17,20,\ldots
]
Progression 2: (s=4), (d=4)
[
4,8,12,16,20,24,28,\ldots
]
This progression contains the perfect squares (4) and (16).
Progression 3: (s=1), (d=5)
[
1,6,11,16,21,26,31,\ldots
]
This progression contains the perfect squares (1) and (16).
Progression 4: (s=0), (d=7)
[
0,7,14,21,28,35,42,\ldots
]
This progression contains the perfect square (0), along with other perfect squares later in the progression.
Progression 5: (s=3), (d=2)
[
3,5,7,9,11,13,15,\ldots
]
This progression contains the perfect square (9).
Progression 6: (s=5), (d=6)
[
5,11,17,23,29,35,41,\ldots
]
Progression 7: (s=6), (d=8)
[
6,14,22,30,38,46,54,\ldots
]
2. A Progression With No Perfect Squares
I predict that some arithmetic progressions can completely avoid perfect squares because of the remainders their terms have when divided by certain numbers.
For example, let (s=2) and (d=4). The progression is
[
2,6,10,14,18,22,26,30,\ldots
]
The formula is
[
x_n=2+4n.
]
Every number in this progression has a remainder of (2) when divided by (4). In other words,
[
x_n\equiv 2\pmod 4.
]
However, a perfect square can only have a remainder of (0) or (1) when divided by (4). For example, the square of an even number is divisible by (4), while the square of an odd number has a remainder of (1).
Since every term in this progression has a remainder of (2), none of the terms can ever be perfect squares.
3. Can a Progression Have Exactly One Perfect Square?
At first, I thought it would be possible to create an arithmetic progression with exactly one perfect square. However, after looking at the pattern more closely, I realized that this is not possible for an infinite arithmetic progression.
Suppose one term in a progression is a perfect square:
[
a^2=s+nd.
]
Now consider the larger square ((a+d)^2). Expanding it gives
[
(a+d)^2=a^2+2ad+d^2.
]
Since (a^2=s+nd), we can substitute it into the equation:
[
(a+d)^2=s+nd+2ad+d^2.
]
Factoring out (d) gives
[
(a+d)^2=s+d(n+2a+d).
]
This has the same form as the original arithmetic progression, so ((a+d)^2) must also appear later in the progression. The same process can be repeated with ((a+2d)^2), ((a+3d)^2), and so on.
Therefore, if an infinite arithmetic progression contains one perfect square, it will actually contain infinitely many perfect squares. This means that a progression can contain either no perfect squares or infinitely many perfect squares, but it cannot contain exactly one.
4. Other Patterns I Noticed
One pattern I noticed is that modular arithmetic is useful for determining whether a progression can contain perfect squares. For example, perfect squares can only have certain remainders when divided by numbers such as (3), (4), (5), or (8). If every term in a progression has a remainder that a perfect square cannot have, then the progression will contain no perfect squares.
I also originally thought that the increasing gaps between perfect squares would eventually stop an arithmetic progression from reaching more squares. However, this is not true because the progression does not need to hit consecutive perfect squares. It can skip many terms before reaching another square. The algebra above shows that once one square appears, other squares will continue appearing later in the progression.
Overall, the main pattern is that the remainder of the seed modulo the constant difference determines whether perfect squares can appear. If one perfect square matches that remainder, then infinitely many other perfect squares will also match it.
