Thread: Find all positive integers for an equation

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

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