For positive integers x and y, x^2 = 350y. Is y divisible

For positive integers x and y, x^2 = 350y. Is y divisible by 28?

Given: \(x^2=2*5^2*7*y\). Notice that since \(2*5^2*7*y\) equals to some perfect square (x^2) then \(y\) must complete the odd powers of other multiples to even numbers (remember perfect square has even powers of its primes), so the least value of \(y\) is \(2*7=14\) (y is a multiple of 14). So, we need one more 2 for \(y\) in order it to be divisible by 28.

(1) x is divisible by 4 --> \(x^2\) is divisible by 4^2=16=2^4, since 350 has only one 2 then \(y\) must have the remaining 2^3, so we have that \(y\) is divisible by 2^3*7=2*28. Sufficient.

(2) x^2 is divisible by 28 --> since the least value of \(y\) is \(2*7=14\) then \(x^2=2*5^2*7*14=28*(5^2*7)\), so we already knew that \(x^2\) is divisible by 28. Not sufficient.

Answer: A.

Hope it's clear.