Suppose a=p(1)^c(1)*...*p(k)^c(k) and b=p(1)^d(1)*...*p(k)^d(k). Prove that if a^3|b^3 then a|b.?

Question

I have used p(1), c(1),d(1), etc. to mean p subscript 1, c subscript 1, and etc. Also, c(1)...c(k) and d(1)...d(k) are non-negative integers. And p(1)...p(k) represent different primes.

Answer 1

Essence of the proof:

1. a divides b if and only if every c(i) is <= d(i)

2. For exactly the same reason, a^3 divides b^3 if and only if every 3c(i) <= 3d(i)

If 2. isn't immediately obvious to you, then work out the prime factorizations of a^3 and b^3.

:)

Answer 2

a^3 - b^3 = sixty one (a - b)(a^2 + ab + b^2) = sixty one if (a - b) = a million, then a^2 + ab + b^2 = sixty one a - b = a million a = b + a million a^2 + ab + b^2 = sixty one (b + a million)^2 + (b + a million)b + b^2 = sixty one ((b + a million)(b + a million)) + b^2 + b + b^2 = sixty one (b^2 + b + b + a million) + 2b^2 + b = sixty one b^2 + 2b + a million + 2b^2 + b = sixty one 3b^2 + 3b + a million = sixty one 3b^2 + 3b - 60 = 0 3(b^2 + b - 20) = 0 3(b + 5)(b - 4) = 0 b = -5 or 4 a = b + a million a = (-5 or 4) + a million a = -4 or 5 a = 5 b = 4 a^2 - b^2 (5)^2 - (4)^2 25 - sixteen 9 ANS : 9 or a = -4 b = -5 (-4)^2 - (-5)^2 = sixteen - 25 = -9 ANS : -9 ANS : -9 or 9 whether you went with b = a million - a, you will nevertheless get the comparable answer.