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About Ramsey2879

  • Birthday May 15

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    Math Cycling
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    George Washington University
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    Retired Patent Examiner
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  • Lepton

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Lepton (1/13)



  1. I think it would be OK to work this out since this post has aged a bit and doing a someone's "homework" doesn't seem to apply anymore. Lets assume the string consists of only 0's and 1's ; that there must be at least a leading 1, and must contain at least one string of at least 4 consecutive zeros. Question: is this what is wanted-The number of strings of length N of this type?
  2. My proof extends to still a more general case that includes Pythagorean Triples Theorem For [math]A,B,x,y,m \in Z | (A,B = Constants)[/math] Let [math] N = F_{x,y} = x^{2} + Bxy -y^{2}[/math] where gcd(x,y) = 1 Then there exists a coprime pair [math](x_{1},y_{1})[/math] such that [math]F_{x_{1},y_{1}} = N^{2^{m}}[/math] I have a recursive formula that gives [math](x_{1},y_{1})[/math] It even gives values if [math] N = F_{x,y} = Ax^{2} + Bxy -y^{2}[/math] e.g. [math](x_{1},y_{1})[/math] such that [math]F = N^{2}[/math] If A is a perfect square and B = 0 then my formula relates to Pythagorean triples e.g. [math]2^{2} - 1^{2} = 3[/math] [math]5^{2} - 4^{2} = 3^{2}[/math] [math]41^{2}-40^{2}=3^{4}[/math] [math]3281^{2}-3280^{2}=3^{8}[/math] [math]3^{2}-2^{2}=5[/math] [math]13^{2}-12^{2}=5^{2}[/math] [math]313^{2}-312^{2}=5^{4}[/math] [math]195313^{2}-195312^{2}=5^{8}[/math]
  3. You can also solve if you substitute a+b for x and set a-1 = -(a+2). This gives a = -1/2. So (-3/2 + b)*(3/2+b) = +/- 3 gives b = +/-sqrt{21/4} since x is real. x = a+b = (-1+/-sqrt{21})/2
  4. I think you need to use the fundamental theorem of arithmetic. i.e. that there is one and only one way to express an integer greater than 1 as a product of primes. (a difference in the order in which the primes are multiplied does not count as a different way). Thus 30 = 2*3*5 = 3*2*5= 5*2*3 etc. There is one 2, one 3, and one 5 in each expression of 30 as a product of primes so that counts as only one way. What does this say about n*n?
  5. Actually my problem is to find my proof that there are an infinite number of solutions to the Diophantine equation 5a^2 + 5ab + b^2 = p^n where a,b are coprime and p is a prime ending in 1 or 9. There is a relationship between any three consecutive term of a Fibonacci type series and the form 5a^2 + 5ab + b^2 that is invariant with the index number of the first term. That is a key to my proof. I leave the proof for you to figure out, but will make suggestions if you reach a dead end and have no idea where to turn. First off then, how does the form 5a^2 + 5ab + b^2 relate to three terms of a Fibonacci type series in an invariant manner? (By Fibonacci type, I mean F(n) = F(n-1) + F(n-2) )
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