Walking the beira mar here in Sao Jose, SC. Keeping my mind busy during the fake virus scare / Bank Robbery. This entire entry is about what I am learning from calculating the squares of the numbers 0 through 100 mentally as I walk the beira mar. First thing you see is that the lowest 2 digits are limited in their values and cyclical.
The last two digits are only unique up to 24. The numbers are:
Last two digits of perfect squares

It is interesting to note that all polynomials are periodic at maximum in the modulus number. However in the above graph you see two full cycles in the span of the modulus of 100. This lead me to investigate the conditions under which 2 cycles occur.
Thus must equal zero
identically
and thus the condition for 2 full cycles in M integers is
<— I have problems reducing these forms of modulus with divisors other than brute force: there is a bit of a gap between here and the next step which needs filled for this procedure to be generalized.
This means the requirement for 2 cycles in M integers is that M must have a factor of 4 in it. The values M=….100,80,60,40…..all have double periodicity.
Setting M=4 yields
Which shows that a perfect square must be:
A natural subsequent question is if 3x periodicity is possible. For that we have
removing the like terms from both sides
The problem here is that if we choose M=90 for instance so the second term's modulus = 0 then the first term is
Which with N the running variable it is not always equal to zero. Thus it appears 3 cycles within the M cyclicity is ruled out. In fact if you look carefully it appears only 2 cycles are possible within the M periodicity because the same term that threw a wrench in the works spoils all others. It appears the reason this works for cyclicity = 2 is that the coefficient of a square polynomial cancels out the fraction in the 1st order term and thus is only available or this cyclicity.
Observations
 Where ever you place a tangent line on the parabola the displacement from the tangent line has the basic parabola ….16,9,4,1,0,1,4,9,16…. around the tangent point. You can see this in the table below and it is used to ease the task of memorization. This is easiest to see around the "lines of symmetry".
On the memorization of the square of 0 – 100
 a=Atom means you have to memorize – you can start by using the (xy)(x+y) trick. Ex: , These are the unique last 2 digits of perfect squares.
 ap= atomic prime – means you can use an atomic to calculate it but soon you will be rattling it off from memory
 Jump == hundreds digit jump
 L.O.S. == Line of Symmetry
Int_  Type  Jump  Square  Step  L.O.S. 
00  a  0000  
01  a  0001  
02  a  0004  
03  a  0009  
04  a  0016  
05  a  0025  
06  a  0036  
07  a  0049  
08  a  0064  
09  a  0081  
10  a  0100  
11  a  0121  
12  a  0144  
13  a  0169  
14  a  0196  
15  a  0225  
16  a  0256  
17  a  0289  
18  a  0324  
19  a  0361  
20  a  0400  
21  a  0441  
22  a  0484  
23  a  0529  
24  a  0576  
25  a  0625  1st  
26  ap  0676  576+100  
27  ap  0729  529+200  
28  ap  0784  484+300  
29  ap  0841  441+400  
30  ap  0900  400+500  
31  ap  0961  361+600  
32  ap  1024  324+700  
33  ap  1089  289+800  
34  ap  1156  256+900  
35  ap  1225  225+1000  
36  ap  1296  196+1100  
37  ap  1369  169+1200  
38  ap  1444  144+1300  
39  ap  1521  121+1400  
40  ap  1600  100+1500  
41  1681  1600+81  
42  100  1764  1700+64  
43  100  1849  1800+49  
44  100  1936  1900+36  
45  100  2025  2000+25  
46  100  2116  2100+16  
47  100  2209  2200+09  
48  100  2304  2300+04  
49  100  2401  2400+01  
50  100  2500  2500+00  2nd  
51  100  2601  2600+01  
52  100  2704  2700+04  
53  100  2809  2800+09  
54  100  2916  2900+16  
55  100  3025  3000+25  
56  100  3136  3100+36  
57  100  3249  3200+49  
58  100  3364  3300+64  
59  100  3481  3400+81  
60  100  3600  3500+100  
61  100  3721  3600+121  
62  100  3844  3700+144  
63  100  3969  3800+169  
64  100  4096  3900+196  
65  200  4225  4000+225  
66  100  4356  4100+256  
67  100  4489  4200+289  
68  200  4624  4300+324  
69  100  4761  4400+361  
70  200  4900  4500+400  
71  100  5041  4600+441  
72  100  5184  4700+484  
73  200  5329  4800+529  
74  100  5476  4900+576  
75  200  5625  5000+625  3rd  
76  100  5776  5100+676  
77  200  5929  5200+729  
78  100  6084  5300+784  
79  200  6261  5400+841  
80  200  6400  5500+900  
81  100  6561  5600+961  
82  200  6724  5700+1024  
83  100  6889  5800+1089  
84  200  7056  5900+1156  
85  200  7225  6000+1225  
86  100  7396  6100+1296  
87  200  7569  6200+1369  
88  200  7744  6300+1444  
89  200  7921  6400+1521  
90  200  8100  6500+1600  
91  200  8281  8200+81  
92  200  8464  8400+64  
93  200  8649  8600+49  
94  200  8836  8800+36  
95  200  9025  9000+25  
96  200  9216  9200+16  
97  200  9409  9400+09  
98  200  9604  9600+04  
99  200  9801  9800+01  
100  200  10000  10000+00 
Graph of
The above graph is the cube of the first 100 integers with the modulus selected to give 3 cycles over the mod 100 cycle. Looking at the form
Seeing this form it is apparent that 3 cycles can be realized within the modulus cycle with the correct choice of modulus. Setting:
Expanding
 The first term is obviously 0 under all conditions.
 The 2nd and 3rd term can be engineered by picking M to make them always zero. The minimum value that works is 9 which you can see by playing with substituting numbers and looking at the resulting terms of the modulus.
Setting M=9 shows:
Which shows that a perfect cube must be:
In the mathematics of sums of powers, it is an open problem to characterize the numbers that can be expressed as a sum of three cubes of integers, allowing both positive and negative cubes in the sum. A necessary condition for {\displaystyle n} to equal such a sum is that {\displaystyle n} cannot equal 4 or 5 modulo 9, because the cubes modulo 9 are 0, 1, and −1, and no three of these numbers can sum to 4 or 5 modulo 9.^{[1]} It is unknown whether this necessary condition is sufficient. See: Wikipedia: Sums of three cubes
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