Another proof of the end two-digit type of the complete square number

1. Question (from Rosen, "Elementary number theory and its application" 6th P99 5th)

The last two decimal digits (bits and 10) that prove the total square number must be one of the following: {xx, E1, E4, O6, E9}

Note: E = even number, O = prime number, 0 is also an even number

2. Verify

N	N2	End number Pair Type
0	0	00
1	1	E1
2	4	E4
3	9	E9
4	16	E4
5	25	25
6	36	O6
7	49	E9
8	64	E4
9	81	E1

3. Proof

3.1 Ideas

To prove the number of bits and 10 digits of the N2, it is simple to express it to see if some rules are met

N2 = n * n

According to our method of calculating multiplication:

Set N2 = AN...A2A1 (an∈{0, 1, ..., 9})

The data placed above the hundred have not affected the bit and 10 bits.

Set F (x) = x bits and 10 digits of the sequence

N2 10-bit and single-digit f (n2) = f (a1a1 + 10*A2A1 + 10*a2a1)

= f (A12 + 10* (2A2A1))

∴n2 digit = A12, n2 10 bits = (2A2A1) + (10 bits of A12)

A1	A12 + 10* (2A2A1)	2A2 (10-bit)	End number Pair Type
0	0	0	00
1	10* (0+2A2) + 1	2A2 = E	E1
2	10* (0+4A2) + 4	4A2 = E	E4
3	10* (0+6A2) + 9	6A2 = E	E9
4	10* (1+8A2) + 6	1+8A2 = O	E4
5	10* (2+10A2) + 5	2+10A2 = 2	25
6	10* (3+12A2) + 6	3+12A2 = O	O6
7	10* (4+14A2) + 9	4+14A2 = E	E9
8	10* (6+16A2) + 4	6+16A2 = E	E4
9	10* (8+18A2) + 1	8+18A2 = E	E1

∴ for all A1, no matter what the value of A2, F (N2) is established, the certificate

4. Inference

According to the 3rd part of the proof, we can also know:

1> as long as the digit of n = 5, then N2 digit and 10 bit = 25, and vice versa is also established

2> as long as the digit of n = 0, then n2 digit and 10 bit = 00, and vice versa is also established

3> as long as the digit of n = 6, then N2 digit and 10 bit = O6, and vice versa is also established

4> regardless of the value of A2, N2 bits and 10-bit sequence and A12, that is, n2 bits and 10-bit series type only and n digits related

Another proof of the end two-digit type of the complete square number