[math-fun] x^6+y^10=z^15
We start by pondering a general problem. Suppose a,b,c are fixed positive integers. Under what conditions does the equation x^a + y^b = z^c have positive integer solutions? -------- If gcd(a,b,c) >= 3, then there are none by Fermat's last theorem. Certain other cases can be eliminated by infinite descent, such as (a,b,c) = (4,4,2). If two of gcd(a,b), gcd(b,c) and gcd(c,a) are 1, then we can find infinitely many solutions by applying the Sam Cappleman-Lynes technique. For example, for x^3 + y^6 = z^7, we start with a solution to u^3 + v^6 = w and multiply throughout by w^6 to give (uw^2)^3 + (vw)^6 = w^7. If gcd(a,b) = gcd(b,c) = gcd(c,a) = 2, then we have infinitely many solutions by applying the Sam Cappleman-Lynes technique to Pythagorean triplets. For example, x^6 + y^10 = z^14 has infinitely many solutions. -------- This does not cover all equations. Does x^6 + y^10 = z^15 have solutions in the positive integers? Mathematica 8 cannot do it, nor can it even solve the simple x^2 + y^3 = z^4, which has a small solution of (x,y,z) = (50625,3375,450) by SCLT. Sincerely, Adam P. Goucher
On Mon, Aug 20, 2012 at 5:25 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
We start by pondering a general problem. Suppose a,b,c are fixed positive integers. Under what conditions does the equation x^a + y^b = z^c have positive integer solutions?
See Beal's Conjecture. http://en.wikipedia.org/wiki/Beal's_conjecture
participants (2)
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Adam P. Goucher -
W. Edwin Clark