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?
Best Answer
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 $4$.