Let a, b and c be natural numbers. If a^2 = bc and gcd(b, c) = 1, then there exist natural numbers s and t such that b = s^2 and c = t^2 .

Question

We were given this proof under our Fundamental Theorem of Arithmetic section and it is supposed to involve prime factorization. I'm totally lost so far and could use any help you can offer.

score 0 · Answer 1 · answered Nov 17 '20 at 00:14

0

Hint.

Write the prime factorization of $bc$.

What can you say about the parity of the power to which each prime is raised?

Can $b$ and $c$ share any of the prime factors of their product?

answered Nov 17 '20 at 00:14

Ethan Bolker

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Let a, b and c be natural numbers. If a^2 = bc and gcd(b, c) = 1, then there exist natural numbers s and t such that b = s^2 and c = t^2 .

1 Answers1