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Old 2007-07-08, 17:27   #3
axn
 
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Jun 2003

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Quote:
Originally Posted by mfgoode View Post
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
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