1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225,... [A000290]

**9 = 4 + (2*(3-1) + 1)**

**16 = 9 + (2*(4-1) + 1)**

**25 = 16 + (2*(5-1) + 1)**

**.**

**.**

**.**

**Claim**: If a

_{2}= 4 then a

_{i}= a

_{i-1}+ (2*(i-1) + 1) where i > 2 generates the sequence formed by squaring integers consecutively.

Proof: Although I never proved this when I first made this claim, the proof of this is very simple and it involves basic algebra.

Since a

_{i}= n

^{2}, a

_{i-1}= (n-1)

^{2}, and (i-1) = (n-1), the a

_{i}= a

_{i-1}+ (2*(i-1) + 1) recurrence relation becomes:

n

^{2}= (n-1)

^{2}+ (2*(n-1) + 1)

= (n

^{2}- 2*n + 1) + (2*n - 2 + 1)

= n

^{2}-

= n

^{2 }

.'.

I used this recurrence relation to find the smallest square root of ((n+1)/2)

^{2}mod n, which I then used to find lonely squares, or integers k such that k

^{2}= 1 mod n where n is the product of distinct primes because I claimed a lonely square is just 2 times the smallest square root of ((n+1)/2)

^{2}mod n.

There are also another type of interesting integers, which I endearingly call

*foursome squares*.

**Definition:**A

*foursome square*is an integer k such that k

^{2}= 4 mod n

Example: Let n = 35, then k = 23 since 23

^{2}= 4 mod 35 but also 12

^{2}= 4 mod 35 and also 2

^{2}= 4 mod 35 and 33

^{2}= 4 mod 35

**Claim:**If n is the product of 2 distinct odd primes and k is a foursome square, then k*((n+1)/2) mod n is a lonely square.