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 Always multiple (Posted on 2008-09-07)
Prove that for all integer values of x,

x9 - 6x7 + 9x5 - 4x3

is an exact multiple of 8,640.

 See The Solution Submitted by pcbouhid Rating: 4.0000 (2 votes)

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 Solution Comment 1 of 1

Let, f(x) = x^9 - 6*x^7 + 9*x^5 - 4*x^3

Then, factorizing f(x), we have:

f(x) = x^3*(x+1)^2*(x-1)^2*(x+2)*(x-2)

If x = 3p, then f(x) is divisible by (3p)^3 = 27*p^3
If x= 3p+1, then f(x) is divisible by (3p)^2*(3p+3) = 27*(p^2)*(p+1)
If x = 3p+2, then f(x) is divisible by (3p+3)^2*(3p) = 27*p*(p+1)^2

Thus, f(x) is always divisible by 27......(i)

Again, f(x) = x^2(x^2 -1)*g(x), where:
g(x) = (x+2)(x+1)x(x-1)(x-2).

Since f(x) is divisible by the product of five consecutive integrs, it follows that f(x) must always be divisible by 5.......(ii)

If x is even, then either x = 4p, or (4p+2)

For x = 4p, f(x) is divisible by (4p)^3*(4), so that f(x) is divisible by 256

For x = 4p+2, f(x) is divisible by 128*(2p+1)^3*p(p+1), so that f(x) is divisible by 128*2 = 256.

Accordingly, f(x) is divisible by 256, whenever f(x) is even.

If x is odd, then x^2-1 is divisible by 8. Let (x^2-1)/8 = M (say). Then, writing:
f(x)= x^3*(x^2-1)^2*(x^2-4), we observe that:

f(x) is divisible by  64*M^2, so that f(x) is divisible by 64, whenever x is odd.

Thus, f(x) is always divisible by 64......(iii)

Combining (i), (ii) and (iii), it follows that in general f(x) is an exact multiple of  LCM(27,5,64) = 8,640.

Edited on September 8, 2008, 6:10 am
 Posted by K Sengupta on 2008-09-07 13:14:50

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