Prove two composite numbers are the same

Q: Consider two composite numbers A and B. Since A is composite we can write:

A = (p1^v1)(p2^v2)...(pn^vn)

Where p(i) is prime and v(i) is an integer. Suppose B is made up of the same primes, the difference is their exponent:

B = (p1^w1)(p2^w2)...(pn^wn) : w(i) is an integer

Show that A=B if and only if v(i) = w(i) for all i.

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A: (Necessary condition) Suppose v(i) = w(i). Clearly A=B.

But how can one deduce v(i) = w(i) given A=B as the starting point. How does one prove the sufficient condition?

My gratitude in advance to those bright sparks that prove this proposition.

Re: Prove two composite numbers are the same

Quote:

Originally Posted by

**bourbaki87** Q: Consider two composite numbers A and B. Since A is composite we can write:

A = (p1^v1)(p2^v2)...(pn^vn)

Where p(i) is prime and v(i) is an integer. Suppose B is made up of the same primes, the difference is their exponent:

B = (p1^w1)(p2^w2)...(pn^wn) : w(i) is an integer

Show that A=B if and only if v(i) = w(i) for all i.

Comment: it usual in composite numbers for $\displaystyle v_i$ to be non-negative integers.

Take a simple case.

GIVEN: $\displaystyle 2^x\cdot 3^y=2^{13}\cdot 3^{14}$.

What must $\displaystyle x~\&~y$ equal and why?

Re: Prove two composite numbers are the same

Answer:

The fundamental theorem of arithmetic: Every integer can be written as a unique product of primes (up to ordering of factors). Thus A=B has only one such product of primes, due to the uniqueness asserted by the F.T.A. Therefore all exponents of primes in A must be equal to the exponents of primes in B thus v(i) = w(i). Q.E.D