Question
If a,b,c,d are four integers such that m = (ab+cd)^2-\dfrac{1}{4}(a^2+b^2-c^2-d^2)^2 is an nonzero integer. Show that |m| must be a composite number
If a,b,c,d are four integers such that m = (ab+cd)^2-\dfrac{1}{4}(a^2+b^2-c^2-d^2)^2 is an nonzero integer. Show that |m| must be a composite number
Using the difference of squares formula,
m = (ab+cd)^2-\dfrac{1}{4}(a^2+b^2-c^2-d^2)^2
=[(ab+cd)+\dfrac{1}{2}(a^2+b^2-c^2-d^2)][(ab+cd)-\dfrac{1}{2}(a^2+b^2-c^2-d^2)]
=\dfrac{1}{4}[(a+b)^2-(c-d)^2][(c+d)^2-(a-b)^2]
=\dfrac{1}{4}(-a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)
Since m is a nonzero integer, m must be divisible by 4.
Therefore, |m| must be a composite number