Expert: Scott A Wilson - 4/30/2009

Question
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Your Question was:

w,x y,z are integers or zero
a,b,c are of 2^x * 3^y * 5^z and have no common divisor
a   +  w =  b =  c -  w ;          for example:
125 - 61 = 64 =  3 + 61
 5 + 11 = 16 = 27 - 11
 5 -  1 =  4 =  3 +  1
 1 +  1 =  2 =  3 -  1
 6 -  1 =  5 =  4 +  1
 8 -  3 =  5 =  2 +  3
32 -  7 = 25 = 18 +  7
48 - 23 = 25 =  2 + 23
 1 +  4 =  5 =  9 -  4
 4 -  1 =  3 =  2 +  1
10 -  1 =  9 =  8 +  1
16 -  7 =  9 =  2 +  7
50 - 23 = 27 =  4 + 23
160 - 79 = 81 =  2 + 79
 1 +  2 =  3 =  5 -  2
Is there  any b greater then 81 ???
I couldn't find anything else below 10^128!!!

Following is the reason:

Hi Hermann~
  What if a = 2^3 * 3^3 * 5^3 = 8*27*125 = 27,000
and w = 0 so be = 27,000 as well as c = 27,000. What is the question? You wanted to know if there is a b greater than 81,and here b is 27,000.  a = 2^2 * 3^2 * 5^1 = 4*9*5 = 180 > 81 but less than 10^128? Remember I used w = 0. I don't understand what it is you want to know??

Math Prof

Expert: Sherry Wallin
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I said:
"a,b,c are of 2^x * 3^y * 5^z and have no common divisor"
and precise
The greatest common divisor of (a,b,c)is 1 and only 1
so if w=0 then a=b=c = the greatest common divisor and hence couldn't be greater then 1
I searche for natural solution too the double angle equation < sin(a)+sin(c) = 2*sin(b)*cos(d) >
A kind of modulation by harmonic frequencys
the general equation would be therefor
sin(2^k*3^l*5^m*theta)+sin(2^r*3^s*5^t*theta) =2*sin(2^n*3^p*5^q*theta)*cos(w*theta)
and cos(0*theta) doesn't make a lot difference as it's equal a factor one and i am only looking for simplified solution ;theta the unite frequency does the rest.
I created Database of all possible frequncies (of  the type 2^x*3^y*5^z*theta)up too 10^129 and tested if any somme of two elements divided by two were a thered member of the Database ; followed by checking if not all three member could not be devided either by 2, 3 or 5.
From their i got the list i send you in the first mail.
But i am not sure that processing after 4*10^9 does have still enough precision to be meanningfull.
Afterwards,i tried to analys the problem and found 5 Subcases :  <w impair>
-  3^l(+-)w   = 2^n = 5^t(-+)w
-  2*2^k(+-)w = 3^p = 2*5^t(-+)w
-  2(+-)w     = 3^p = 2^r*5^t(-+)w
-  2*2^k(+-)w = 5^q = 2*3^s(-+)w
-  2(+-)w     = 5^q = 2^r*3^s(-+)w
and 4 further subcase: < w is pair>
-  3^l*5^m(+-)w = 5^q     = 3^s(-+)w
-  5^m(+-)w     = 3^p*5^q = 3^s(-+)w
-  5^m(+-)w     = 3^p     = 3^s*5^t(-+)w
-  1(+-)w       = 3^p*5^q = 3^s*5^t(-+)w
and least i analysed and factoriesed the table of 5^z-3^y
and found one interesting relation of ([5^(2^h)]^2-[3^(2^h)]^2)/(2*[5^(2^h)-3^(2^h)]) =([5^(2^h)+3^(2^h)])/2
which give the smallest noncommon factors in respect to 2,3,5 for Great Numbers. and even soo there is un exponential growth, but does this anything prouve.
i am sure there is a relation between the exponents and the valeur of w.<a litle bit like in fermat's last problem>
besides this, the accummulation of probabilite does after the expantion out of the local region nevermore trespass unite. but does a probabilty smaller than 1 means that their is no event at all? all this is no excact mathematics!

i am not sure that i am any more clear,
but would be fine if you could look into it a seconde time.

Have some fun
Hermann  

Answer
Someone suggested that if you take the sum of a_i to the n,
i needs to go from 1 to n.  In other words, a^2+b^2=c^2,
a^3+b^3+c^3=d^3, a^4+b^4+c^4+d^4=e^4, etc.

They have found this breaks down at 5.  In other words,
if you take 4 numbers to the fifth power,
you can get another number to the fifth power.

The number are quite large and most computer can barely do them
when they are taken to the 4th.  From this, I can say that there is certainly no way powers of 2,3, and 5 would work.

Note that for large powers, the difference between two number that only differ by 1 is quite large.  The difference between (n+1)^4 and n^4 is 3n^2 + 3n + 1.  That is certainly some number greater than another small number to the 4th.

If you look at (n+1)^6 - n^6 you get 6n^5+15n^4+20n^3+15n^2+6n+1.

Notice I looked at number that only differed by 1.  If they had much bigger difference, the difference would be a whole lot larger.
Eventually when the numbers get so large, the difference begins to get under control, but there is no way it can be done with 2, 3, and 5.