2021 WSMO Accuracy Round Problems/Problem 3

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Problem

If $f$ is a monic polynomial of minimal degree with rational coefficients satisfying $f\left(3+\sqrt{5}\right)=0$ and $f\left(4-\sqrt{7}\right)=0,$ find the value of $|f(1)|$.

Solution

If $3+\sqrt{5}$ is a root of the polynomial, then $3-\sqrt{5}$ is too. Similarly, if $4-\sqrt{7}$ is a root of the polynomial, then so is $4+\sqrt{7}.$ Thus, $f(x)=d(x)\left(x-\left(3+\sqrt{5}\right)\right)\left(x-\left(3-\sqrt{5}\right)\right)\left(x-\left(4-\sqrt{7}\right)\right)\left(x-\left(4+\sqrt{7}\right)\right),$ for some polynomial $d.$ To minimize the degree of $f,$ we can set the degree of $d$ to 0, which means that $d(x)=c.$ This means that the leading coefficient of $f(x)=c.$ Since $f$ is monic, $c=1,$ which means that \[f=\left(x-\left(3+\sqrt{5}\right)\right)\left(x-\left(3-\sqrt{5}\right)\right)\left(x-\left(4-\sqrt{7}\right)\right)\left(x-\left(4+\sqrt{7}\right)\right)=\]\[\left(\left(x-3\right)-\sqrt{5}\right)\left(\left(x-3\right)+\sqrt{5}\right)\left(\left(x-4\right)+\sqrt{7}\right)\left(\left(x-4\right)-\sqrt{7}\right)=\]\[\left(\left(x-3\right)^{2}-\left(\sqrt{5}\right)^{2}\right)\left(\left(x-4\right)^{2}-\left(\sqrt{7}\right)^{2}\right)=\]\[\left(x^{2}-6x+4\right)\left(x^{2}-8x+9\right)=x^{4}-14x^{3}+61x^{2}-86x+36.\] We conclude that \[|f(1)|=|1-14+61-86+36|=|-2|=\boxed{2}.\]

~pinkpig