### 2001 IMO SL NT #5

Let $a > b > c > d$ be positive integers and suppose that$ac + bd = (b+d+a-c)(b+d-a+c).$Prove that $ab + cd$ is not prime.

Solution 1 (completely did not expect this to be geo, extremely elegant)

We are given

$$ab+cd=(b+d)^2-(a-c)^2$$

$$=b^2+2bd+d^2-a^2+2ac-c^2$$

$$a^2-ac+c^2=b^2+bd+d^2.$$

Let us construct a quadrilateral $WXYZ$ such that $WX=a, XY=c, YZ=b, ZW=d,$ and $WY=\sqrt{a^2-ac+c^2}=\sqrt{b^2-bd+d^2}.$

Then by the law of cosines, we obtain $∠WXY = 60^\circ$and $∠WZY = 120^\circ$. Hence this quadrilateral is cyclic. By Strong Ptolemy, we have $$WY^2=\frac{(ab+cd)(ad+bc)}{(ac+bd)}.$$

Now assume for the sake of contradiction that $ab+cd$ is a prime $p$. Since $a>b>c>d$, by the rearrangement inequality we have $$p=ab+cd>ac+bd>ad+bc.$$ Let $y=ac+bd$ and $x=ad+bc$ now. The point is that $$p \cdot \frac{x}{y}$$cannot possibly be an integer if $p$ is prime and if $x<y<p$. But $WY^2=a^2-ac+c^2$ is clearly an integer as $a,c$ are integers themselves, so this is a contradiction and $p=ab+cd$ cannot be a prime. $\square$

Solution 2 (standard)

We see $$b+d+a-c|ac+bd \implies b+d+a-c|ac+bd+ab+ad+a^2-ac=(a+b)(a+d).$$

Similarly, we see that $$b+d-a+c|(c+b)(c+d)$$.

Multiplying the equations gives

$$(b+d+a-c)(b+d-a+c)=ac+bd|(a+b)(a+d)(c+b)(c+d)$$

$$ac+bd|(ac+bd+ad+bc)(ac+bd+ab+cd)$$

$$\implies ac+bd|(ad+bc)(ab+cd).$$Again,

$a(c-d)>b(c-d)(a>b,c>d) \implies ac+bd>ad+bc$

and $a(b-c)>d(b-c)(a>d,b>c) \implies ab+cd>ac+bd>ad+bc$.

This implies as $ac+bd \not | ad+bc \implies \gcd(ac+bd,ab+cd)>1$,

and thus we're done $\square$.