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- Feb 14, 2012

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- Feb 14, 2012

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- Feb 14, 2012

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Difference of squares

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- Feb 14, 2012

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$(x-x_1^2)(x-x_2^2)(x-x_3^2)(x-x_4^2)\\=(\sqrt{x}-x_1)(\sqrt{x}-x_2)(\sqrt{x}-x_3)(\sqrt{x}-x_4)(\sqrt{x}+x_1)(\sqrt{x}+x_2)(\sqrt{x}+x_3)(\sqrt{x}+x_4)\\=P(\sqrt{x})P(-\sqrt{x})$

Let's try this on the given expression.

$(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_1^4+1)\\=(x_1^2-i^2)(x_2^2-i^2)(x_3^2-i^2)(x_1^4-i^2)\\=(x_1-i)(x_2-i)(x_3-i)(x_4-i)(x_1+i)(x_2+i)(x_3+i)(x_4+i)\\=P(i)P(-i)\\=(1-ai-b+ci+d)(1+ai-b-ci+d)\\=(1-b+d+(c-a)i)(1-b+d-(c-a)i)\\=(b-d-1)^2+(c-a)^2$

Now it is clear where the condition $b-d\ge 5$ comes in, we have

$(b-d-1)^2+(c-a)^2\ge (5-1)^2+0^2=16$

But we are not done yet, we need to exhibit a polynomial $P(x)$ that achieves the value 16. For this, we need $b-d=5$ and $c-a=0$ to hold. Fortunately there is any easy polynomial that satisfies this condition:

$(x+1)^4=x^4+4x^3+6x^2+4x+1$ and we conclude that the answer is indeed 16.