Let a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)^7 .

Answer 1

(a+b+c)^7 = (a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)

will be of the form

a^7 + c1*a^6*b + c2*a^6*c + c3*a^5*b^2 + c4*a^5*c^2 + c5*a^5*b*c + . . . .

where c1, c2, c3 are constants found when doing the actual expansion.

so all the terms in the expansion are of the form

ci*a^x*b^y*c^z

where x, y, z are integers from 0 to 7, and x+y+z = 7

So (ci*a^x*b^y*c^z)/abc = ci*a^(x-1)*b^(y-1)*c^(z-1)

So in the case when x, y, and z, are all 1 or greater, abc divides these terms in the expansion.

We are left with the terms when x, y, or z is 0

the terms involving

a^7, b^7, c^7, a^6*b, a^6*c, a*b^6, a*c^6, b^6*c, 6*c^6

a^7 / abc = a^6/bc; but c divides a^2; so we need to show that a^4/b is a real number; b divides c^2, so c^2 is a multiple of b; a^4/b if and only if a^4/c^2; c divides a^2; so c^2 divides a^4; a^7 divides abc

Use this type of logic for the rest of the terms b^7, . . .etc