Find all possible integers n for which the equation X^3 + Y^3 = n!+4 has solutions in integers. I suppose the following: X+Y = (n!+4)^(1/3) Can you help finish this problem? Thanks in advance
Follow Math Help Forum on Facebook and Google+
Originally Posted by fernipascual Find all possible integers n for which the equation X^3 + Y^3 = n!+4 has solutions in integers. Originally Posted by fernipascual I suppose the following: X+Y = (n!+4)^(1/3) These two are not at all same. All you can do is to get the values of from the equation: by taking different values of .
Originally Posted by fernipascual Find all possible integers n for which the equation X^3 + Y^3 = n!+4 has solutions in integers. I suppose the following: X+Y = (n!+4)^(1/3) Can you help finish this problem? Thanks in advance As an addendum to Sambit's response, please note that is not X + Y any more than is equal to X + Y. -Dan
View Tag Cloud