# Problem Of The Week #373 Jul 4th, 2019

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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Let $a, \, b,\,c$ and $d$ be integers with $a>b>c>d>0$.

Suppose that $ac+bd=(b+d+a-c)(b+d-a+c)$.

Prove that $ab+cd$ is not prime.

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#### anemone

##### MHB POTW Director
Staff member
Congratulations to Olinguito for his correct solution, which you can find below!

We have
$$ac+bd=(b+d+a-c)(b+d-a+c)=(b+d)^2-(a-c)^2=b^2+d^2-a^2-c^2+2(ac+bd)$$
$\implies\ a^2-ac+c^2\ =\ b^2+bd+d^2$.

Hence
$$\begin{array}{rcl}(ac+bd)(b^2+bd+d^2) &=& ac(b^2+bd+d^2)+bd(b^2+bd+d^2) \\ {} &=& ac(b^2+bd+d^2)+bd(a^2-ac+c^2) \\ {} &=& ab^2c+acd^2+a^2bd+bc^2d \\ {} &=& (ab+cd)(ad+bc)\end{array}$$
$\implies\ ac+bd\mid(ab+cd)(ad+bc)\ \ldots\ \boxed1$.

But $a>b$ and $c>d$ $\implies$ $(a-b)(c-d)>0$ $\implies$ $ac+bd>ad+bc\ \ldots\ \boxed2$.

Similarly $a>d$ and $b>c$ $\implies$ $(a-d)(b-c)>0$ $\implies$ $ab+cd>ac+bd\ \ldots\ \boxed3$.

If $ab+cd$ were prime, then $\boxed3$ would imply $\gcd(ab+cd,ac+bd)=1$ and then $\boxed1$ would imply $ac+bd\mid ad+bc$, contradicting $\boxed2$. It follows that $ab+cd$ cannot be prime.

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