I have noticed two patterns with square numbers.
$1^2\equiv 1\pmod{10}$
$2^2\equiv 4\pmod{10}$
$3^2\equiv 9\pmod{10}$
$4^2\equiv 6\pmod{10}$
$5^2\equiv 5\pmod{10}$
$6^2\equiv 6\pmod{10}$
$7^2\equiv 9\pmod{10}$
$8^2\equiv 4\pmod{10}$
$9^2\equiv 1\pmod{10}$
This set of numbers ${1, 4, 9, 6, 5, 6, 4, 1}$ is palindromic (meaning it is the same backwards and forwards) and it rotates around $5$. In fact, it rotates around all odd multiples of $5$. And the set of ones digits of squares $6, 9, 4, 1, 0, 1, 4, 9, 6$ rotates around all even multiples of $5$.
That is the first pattern. If $|x-5|=|y-5|,$ then $x^2\equiv{y^2}\pmod{10}$.
The second pattern is similar, except it is bigger. If $|x-25|=|y-25|,$ then $x^2\equiv{y^2}\pmod{100}$.
My questions are: Why do these patterns work? Are the reasons related? And does this pattern appear with any other numbers?
The single digit symmetry can be understood from the fact that $(-1)^2=1$. $8 \equiv -2 \pmod {10},$ so $8^2 \equiv (-2)^2 \equiv 2^2 \pmod {10}$ and similarly for the others. The symmetry around $0$ will be true in any base. The symmetry around the midpoint will be true in any even base because $(5-a)^2=25-10a+a^2, (5+a)^2=25+10a+a^2$ and $+10a \equiv -10a \equiv 0 \pmod {10}$.
The previous paragraph will prove the two digit case is symmetric around $50$, but is not sufficient to prove it is symmetric around $25$. For that, write $(25+a)^2=625+50a+a^2, (25-a)^2=625-50a+a^2$ and $50a \equiv -50a \pmod {100}$. This pattern will work for any base that is a multiple of $4$.