Products of consecutive Integers
... Here the right hand side is a polynomial of u, missed summands have less degree of u, and leading term mukm−k is positive. Hence this polynomial is positive if u is large. So the inequality we need is obtained. 3.2. Follows from the next point. 3.3. It is enough to prove that if m > 2k the inequalit ...
... Here the right hand side is a polynomial of u, missed summands have less degree of u, and leading term mukm−k is positive. Hence this polynomial is positive if u is large. So the inequality we need is obtained. 3.2. Follows from the next point. 3.3. It is enough to prove that if m > 2k the inequalit ...
Modular forms and Diophantine questions
... have little trouble guessing the general formula for the number of solutions to x2 + y 2 = 1 mod p. When p = 2, there are the two solutions (0, 1) and (1, 0). When p is bigger than 2, there are either p + 1 or p − 1 solutions to x2 + y 2 = 1 mod p, depending on whether p is 1 less than or 1 more tha ...
... have little trouble guessing the general formula for the number of solutions to x2 + y 2 = 1 mod p. When p = 2, there are the two solutions (0, 1) and (1, 0). When p is bigger than 2, there are either p + 1 or p − 1 solutions to x2 + y 2 = 1 mod p, depending on whether p is 1 less than or 1 more tha ...
THE INSOLUBILITY OF CLASSES OF DIOPHANTINE EQUATIONS
... infer that there exists no solution of X,m + X,m + X,m e 0 (mod p) and X,X .X, i~6 0 (mod p) . Hence, if there exists a rational solution X,11d + X2+ X,m - 0, then p I X,X.X, . If q denotes any prime factor of 9n, and (X,X2X,, in) =1 we have, by using Furtwangler's criterion on Fermat's Equation (cf ...
... infer that there exists no solution of X,m + X,m + X,m e 0 (mod p) and X,X .X, i~6 0 (mod p) . Hence, if there exists a rational solution X,11d + X2+ X,m - 0, then p I X,X.X, . If q denotes any prime factor of 9n, and (X,X2X,, in) =1 we have, by using Furtwangler's criterion on Fermat's Equation (cf ...
