if (s,m)=1, denote the solution to sx=r, mod m by r/s, prove (r/s)(t/u)=(r/s)(t/
ID: 3082504 • Letter: I
Question
if (s,m)=1, denote the solution to sx=r, mod m by r/s, prove (r/s)(t/u)=(r/s)(t/u), mod m and prove ((r/s)+(t/u))=((r/s)+(t/u)), mod mExplanation / Answer
Starting from x^(1/n)+y^(1/n)=z^(1/n), divide by x^(1/n). Set u = (y/x)^1/n and v = (z/x)^1/n. Then 1+u=v. It is enough to show that u is rational. Indeed if u = r/s with r,s coprime, then y/x = (r/s)^n hence y = cr^n, x = cs^n and z = c(r+s)^n. So we want to show that if u>0 is such that u^n and (1+u)^n are rational, then u is rational. For that we shall show that U = (u,u^2,.....,u^(n-1)) is a set of solutions of a Cramer system with rational entries namely MU^T = V where is a square matrix with rationl entries and non zero determinant and V rational. To obtain the lines of M expand (1+u)^n. Call A = u^n, B= (1+u)^n and C = (1+u)^n - u^n - 1. We denote {k} = (n choose k) for all 0 sum |cj*a(i,j)| and hence sum cj Yj can't vanish. This lemma clearly applies to the previous matrix since for instance -C u + {1}u^2 + 2u^{3}+ ........... +{n-2}u^{n-1}= -{n-1}A implies | -C u | > {1}u^2 + 2u^{3}+ ........... +{n-2}u^{n-1} and same for all lines. Now we can in each column erase the powers u,u^2,...,u^(n-1) and we still have an invertible matrix. This is the matrix of a system whose solutions are (u,u^2,.....,u^(n-1)). Since the right side has rational entries, as well as the matrix, u is rational.Related Questions
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