If y is a positive integer, is (y^4+7)^2 divisible by 4?

If a number is divisible by 4 then it will be an even number.

(1) √y has exactly three Prime Factors

\(\sqrt y\) is square of a prime, as only square of primes have total 3 factors.

If \(\sqrt y =4 \) then \(y = 16 \)

In this case \((y^4+7)^2\) turns out be odd and is NOT divisble by \(4 \)

If \(\sqrt y =9 \) then \(y = 81 \)

\((81^4+7)^2 = (3^{16}+7)^2\) . Little calculation shows that this is divisble by \(4 \)

INSUFF.

(2) All the prime Factors of \(y^7 \) are greater than \(3\)

All prime numbers greater than \(3\) are odd,So \(y\) will be odd and the expression \((y^4+7)^2 \) will be of the form:

\((3^4+7)^2 =\hspace{2mm}\)Divisible by \(4 \)
\((5^4+7)^2 =\hspace{2mm}\)Divisible by \(4\)
\((7^4 +7)^2=\hspace{2mm}\)Divisible by \(4\)
\((11^4 +7)^2=\)Divisible by \(4\)

etc.

Note: One needs to only calculate the last two digits of the above expression to determine whether it's divisble by \(4.\)

Hence we can definitely answer yes to the question

SUFF.

Ans B

Hope it's clear