1963 TMTA High School Algebra I Contest Problem 14

Problem

$16x^4 - y^4$ factored into its real prime factors is equal to:

$\text{(A)} \quad (4x^2+y^2)(2x-y)^2 \quad \text{(B)} \quad (4x^2+y^2)(4x^2-y)^2 \quad \text{(C)} \quad (4x^2+y^2)(2x-y)(2x+y)$

$\text{(D)} \quad 16(x^2+y^2)(x-y)(x+y) \quad \text{(E) can't be factored}$

We can use difference of squares. $16x^4 - y^4 = (4x^2+y^2)(4x^2-y^2) = (4x^2+y^2)(2x+y)(2x-y)$ The answer is $\boxed{\text{(C)}}.$

1963 TMTA High School Mathematics Contests (Problems)
Preceded by Problem 13	TMTA High School Mathematics Contest Past Problems/Solutions	Followed by Problem 15