Jun 2003
2×3^{2}×293 Posts

Quote:
Originally Posted by mfgoode
Show that (x^5)/5 + (x^3)/3 +(7x/15) is always an integer for integral values of x

Alternately we need to show that 3x^5 + 5x^3 + 7x is a multiple of 15.
Working modulo 3, we have x^3=x (by Fermat's Little Theorem)
Thus 3x^5 + 5x^3 + 7x == 3x^5 + 5x + 7x == 3x^5 + 12x == 0 (mod 3)
Working modulo 5, we have x^5=x
Thus 3x^5 + 5x^3 + 7x == 3x + 5x^3 + 7x == 10x + 5x^3 == 0 (mod 5)
Thus our expression is divisible by 15, as required.
Q.E.D
