If a and b are natural number with a^2 divides b^2, then we prove a divides b. W
ID: 1887977 • Letter: I
Question
If a and b are natural number with a^2 divides b^2, then we prove a divides b.We can write b^2=k*a^2 for some integer k.
Since a^2 and b^2 are perfect squares, prove that k must also be a perfect square using the fundamental theorem of arithmetic.
Explanation / Answer
Clearly if each ri is even then there exist integers si>=0 such that ri = 2si for all i. Therefore n = (p1^r1)(p2^r2)...(pk^rk) = (p1^(2s1))(p2^(2s2))...(pk^(2sk)) = [(p1^s1)(p2^s2)...(pk^sk)] ^ 2 so n is a perfect square. If n is a perfect square, say 'a', then if 'a' has the unique factorization (as per the fundamental theorem of arithmetic): a = (q1^s1)(q2^s2)...(qm^sm) where q1,q2,...,qm are distinct primes and each sj>=0, and a^2=n, then a^2 = [(q1^s1)(q2^s2)...(qm^sm) ]^2 = (p1^r1)(p2^r2)...(pk^rk) = n (q1^(2s1))(q2^(2s2)...(qm^(2sm)) = (p1^r1)(p2^r2)...(pk^rk) Since by the fundamental theorem of arithmetic the factorization is unique, m=k and every qj = pi for some j and i. and also every ri = 2sj . Thus every ri is even since it has the factor 2.