Automorphic numbers are numbers of "n" digits whose squares end in the number itself.

Such numbers must end in 1, 5, or 6 as these are the only numbers whose products produce 1, 5, or 6 in the units place. For instance, the square of 1 is 1; the square of 5 is 25; the square of 6 is 36.

What about 2 digit numbers ending in 1, 5, or 6? It is well known that all 2 digit numbers ending in 5 result in a number ending in 25 making 25 a 2 digit automorphic number with a square of 625. No other 2 digit numbers ending in 5 will produce an automorphic number.

Is there a 2 digit automorphic number ending in 1? We know that the product of 10A + 1 and 10A + 1 is 100A2 + 20A + 1. "A" must be a number such that 20A produces a number whose tens digit is equal to "A". For "A" = 2, 2 x 20 = 40 and 4 is not 2. For "A" = 3, 3 x 20 = 60 and 6 is not 3. Continuing in this fashion, we find no 2 digit automorphic number ending in 1.

Is there a 2 digit automorphic number ending in 6? Again, we know that the product of 10A + 6 and 10A + 6 is 100A2 + 120A + 36. "A" must be a number such that 120A produces a number whose tens digit added to 3 equals "A". For "A" = 2, 2 x 120 = 240 and 4 + 3 = 7 which is not 2. For "A" = 3, 3 x 120 = 360 and 6 + 3 = 9 which is not 3. Continuing in this manner through A = 9, for "A" = 7, we obtain 7 x 120 = 840 and 4 + 3 = 7 = "A" making 76 the only other 2 digit automorphic number whose square is 5776.

By the same process, it can be shown that the squares of every number ending in 625 or 376 will end in 625 or 376.

The sequence of squares ending in 25 are 25, 225, 625, 1225, 2025, 3025, etc. The nth square number ending in 25 can be derived directly from N(n)2 = 100n(n - 1) + 25. (This expression derives from the Finite Difference Series of the squares.)