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.   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 (x-y)(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. It is unknown whether this necessary condition is sufficient.   See: Wikipedia: Sums of three cubes

Categories: Math