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

Integer x^2 Mod(100) Integer
00 00 50
01 01 49
02 04 48
03 09 47
04 16 46
05 25 45
06 36 44
07 49 43
08 64 42
09 81 41
10 00 40
11 21 39
12 44 38
13 69 37
14 96 36
15 25 35
16 56 34
17 89 33
18 24 32
19 61 31
20 00 30
21 41 29
22 84 28
23 29 27
24 76 26
25 25  

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. 

N^2 Mod M = [ N - M/2 ]^2 Mod M                       

N^2 Mod M = [ N^2 - MN + M^2 / 2^2 ] Mod M     

N^2 Mod M = N^2 Mod M  - (MN) Mod M + (M^2 / 2^2) Mod M       

Thus    - (MN) Mod M + (M^2 / 2^2) Mod M     must equal zero

   - (MN) Mod M ==0   identically 

and thus the condition for 2 full cycles in M integers is   

 (M^2 / 2^2) Mod M =0       <— 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:  N^2 Mod 4 = 0,1                     


A natural subsequent question is if 3x periodicity is possible. For that we have

N^2 Mod M = [ N - M/3 ]^2 Mod M                       

N^2 Mod M = [ N^2 - (2/3)MN + M^2 / 3^2 ] Mod M        removing the like terms from both sides

0 = [ - (2/3)MN + M^2 / 3^2 ] Mod M         The problem here is that if we choose M=90 for instance so the second term's modulus = 0 then the first term is 

0 = ( - 60N ) Mod M      

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 (x-y)(x+y) trick. Ex:    53^2 = (53-3)(53+3) + 3^2 = 50*56 + 9 =2809   ,   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  N^3 Mod 90   

 

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

    [ N + M ]^3 = N^3 + 3N^2 M + 3N M^2 + N^3     

Seeing this form it is apparent that 3 cycles can be realized within the modulus cycle with the correct choice of modulus.  Setting:

  N^3 Mod M = [ N + M/3 ]^3 Mod M                       

Expanding

  N^3 Mod M =  [ N^3 + 3N^2 M + 3N M^2/3^2 + M^3/3^3 ] Mod M          

  N^3 Mod M =  (N^3)Mod M  + (3N^2 M) Mod M  + (3N M^2/3^2)Mod M  + (M^3/3^3) Mod M          

  0 =  + (3N^2 M) Mod M  + (N M^2/3)Mod M  + (M^3/3^3) Mod M      

  0 =  + (3N^2 M) Mod M  + (N M^2/3)Mod M  + (M^3/3^3) Mod M      

  • The first term is obviously 0 under all conditions.

  0 = (N M^2/3)Mod M  + (M^3/3^3) Mod M      

  • 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:  N^3 Mod 9 = 0,1 ,8      ( -1,0,1)?               

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

 

 

 

 

 

 

 

 

 

 

 

 

 

Categories: Math

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