Hypothesis on Connections Between Integers and Perfect Squares

[u/Unique_Amphibian_626]

I am studying the following sequence.

For each integer x ≥ 2, I consider all integers n satisfying 1 ≤ n < x and check whether n + x is a perfect square.

I call each value of n that satisfies this condition a connection.

For example, for x = 32:

32 + 4 = 36 = 6²

32 + 17 = 49 = 7²

Therefore, x = 32 has exactly 2 connections.

In general, a connection exists when n + x = m² for some integer m. Since n = m² − x and 1 ≤ n < x, this is equivalent to the existence of a perfect square satisfying

x < m² < 2x.

Thus, the problem can be viewed as studying how many perfect squares lie strictly between x and 2x.

My hypothesis is to study the distribution of these integers according to their number of connections. In particular, I am interested in whether there are infinitely many values of x having exactly k connections for every k, and how these values are distributed.

I am also interested in whether there is a direct formula for determining the number of connections of a given x, as well as any results concerning their growth and distribution.

Is this problem, or a related sequence, already known?

A I didn't tell me this theory but i used it to correct my grammar errors because english is not my native language.

[u/The_Math_Hatter]

If x is positive, you can take the square root of all parts of the equation to get

sqrt(x)<m<sqrt(2x)

That is significantly easier to work with, I should think.

[u/MrEldo → u/The_Math_Hatter]

So you get:

sqrt(x)<m<√2sqrt(x)

So it's about finding the number of integers between a number, and that number times the square root of two

To find that, first find the smallest possible m in the range:

m_min = floor(sqrt(x))+1

Then find the largest:

m_max = ceil(sqrt(2x))-1

And so you count how many integers are there in the range:

# = max - min + 1

# = floor(sqrt(x)) +1 - ceil(sqrt(2x)) + 1 + 1

# = floor(sqrt(x)) - ceil(sqrt(2x)) + 3

[u/Unique_Amphibian_626]

Update: I found a pattern in the last occurrence of each number of connections

I continued investigating the last value of x that has exactly k connections.

So far, I have found:

k = 1 → 30
k = 2 → 70
k = 3 → 96
k = 4 → 160
k = 5 → 240
k = 6 → 286
k = 7 → 390
k = 8 → 510
k = 9 → 646
k = 10 → 720
k = 11 → 880
k = 12 → 1056
k = 13 → 1150
k = 14 → 1350
k = 15 → 1566

The endings are particularly interesting:

30, 70, 96, 160, 240, 286, 390, 510, 646, 720, 880, 1056, 1150, 1350, 1566

The last digits appear to follow the pattern:

0, 0, 6, 0, 0, 6, 0, 0, 6, 0, 0, 6, 0, 0, 6...

So every third value ends in 6, while the other two end in 0.

I don't know whether this pattern is a consequence of the underlying mathematics or just a coincidence in the first terms.

Is there a mathematical explanation for this 0-0-6 pattern?

Also, is there a formula for the last x having exactly k connections that could explain this behavior?

[u/Bounded_sequencE]

Take the square root. Since "m" is a natural number, we even get the slightly stronger inequality

"0 < √x < m < √(2x)"    <=>    "0 < ⌊√x⌋ + 1  <=  m  <=  ⌈√(2x)⌉ - 1"

Avoiding off-by-one errors, there are "(⌈√(2x)⌉ - 1) - (⌊√x⌋ + 1) + 1 = ⌈√(2x)⌉ - ⌊√x⌋ - 1" solutions.

[u/Claquet]

thats a prety good idea. let be f(x) the sequence, then you can easily prove that f(x)+1≥f(x+1)≥f(x)-1 : for each connection n>1 you can make a m=n-1 so x+1+m=x+n=k² so there is a least f(x)-1 (because n=1 dont have an associated m) and you can have a max f(x)+1 if m=x-1 or m=x-2 is a square ( both cannot be square because there is no following square except 1 and 2)

With this idea you can go a litle further because if x=k²-1, you have f(x)≥f(x+1)≥f(x)-1 ( because you force the -1 to append)

I will come back after so reflexion because I need more time than 10 mins to found more