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 $\displaystyle Y$ from the equation: $\displaystyle Y = ( n!+4 - X^3)^{1/3}$ by taking different values of $\displaystyle X$.
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 $\displaystyle \sqrt[3]{X^3 + Y^3}$ is not X + Y any more than $\displaystyle \sqrt{X^2 + Y^2}$ is equal to X + Y.