Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Solution by Iura: Lemma. If $ab=cd$ then there exist $m,n,p,q$ with $a=mn, b=pq, c=mp, d=nq$. Proof is obvious. Opening the brackets we get $a^2+c^2-ac=b^2+bd+d^2$ Multiplying by $4$ we get $(2a-c)^2+3c^2=(2b+d)^2+3d^2$ thus $3(c+d)(c-d)=(2b+d-2a+c)(2b+d+2a+c)$. Now using this Lemma we get either \[c+d=mn, c-d=pq, 2a+c-2b+d=3mp, 2b+d+2a+c=nq,\] or \[c+d=mn, c-d=pq, 2a+c-2b+d=mp, 2b+d+2a+c=3nq.\] For the first case, notice that $16(ab+cd)=(3p^2+n^2)(3m-q)(m+q)$, and it's easy to handle it now. The second case is analogous.

this is the solution i found.it is in the same spirit as the above one but a bit different.let$a=x+y+z+d ,b=x+y+d,c=d+x$.using this it is easy to see that the given relation is equivalent with $xy=(z-d)(2d+2y+x+z)$and $ab+cd=(2d+2y+x+z)(d+x)+y(y+z)$.but it is clear that $gcd(2d+2y+x+z,y)=t>1$and it is also clear that $t|ab+cd$so our number is not prime and the problem is over.

I'm aware that there exists a solution using the Eisenstein integers. Could someone post it?

a>b>c>d>0 ac+bd=(b+d+a-c)(b+d+c-a) (*) suppose that ab+cd -prime (*) hence (b+d+a-c)l(ac+bd) we have (b+d+a-c) l (a)(b+d+a-c)+(ac+bd)=ab+ad+a^2+bd=(a+b)(a+d) (**) (*) hence (b+d+c-a) l (ac+bd) we have (b+d+c-a) l (b+d+c-a)(c)+(ac+bd)=cb+cd+c^2+bd=(c+b)(c+d) (***) (**) (***) hence (b+d+c-a)(b+d+a-c) l [(a+b)(c+d)][(a+d)(b+c)] at this we have (ac+bd) l ((ac+bd)+(ad+bc))((ac+bd)+(ab+cd)) (****) (a-d)(b-c)>0 therefore ab+cd>ac+bd and (ab+cd)-is prime hence ((ac+bd);((ac+bd)+(ad+bc)))=1 (*****) (****) and (*****) we can say (ac+bd) l (ac+bd)+(ad+bc) at this we have (ac+bd) l (ad+bc) (ac+bd)<=(ad+bc) it is equalent (a-b)(d-c)>0 contradiction! ab+cd - is not prime

O very nice solutions ALISHER

I was trying the problem in a different approach but could not complete it. Here is my mode of approach. Given that a>b>c>d and (ac+bd)=(b+d+a-c)(b+d-a+c) or, ac+bd=(b+d)^2-(a-c)^2 or, ac+bd=b^2+d^2+2bd-a^2-c^2+2ac or, ac+bd=a^2+c^2-b^2-d^2=(a+b)(a-b)+(c+d)(c-d)=(a+d)(a-d)-(b+c)(b-c) Now, I was trying to use the identity, (ab+cd)+(ac+bd)=(b+c)(a+d) and (ab+cd)-(ac+bd)=(a-d)(b-c) which gives 2[(a+b)(a-b)+(c+d)(c-d)]=(b+c)(a+d)+(a-d)(b-c)........

weird solution: we have: $ ac+bd=(b+d+a-c)(b+d-a+c)\iff\boxed{a^2-ac+c^2=b^2+bd+d^2}$ (*) now let $ ABCD$ be a quadrilateral with $ AB=a,BC=d,CD=b,AD=c$ and $ \angle BAD=60^\circ$ and $ \angle BCD=120^\circ$.now note that according to (*) and $ \cos$ law there exists such quadrilateral because each side of (*) is equal to $ BD^2$. now let $ \angle ABC=\alpha$ hence $ \angle CDA=180^\circ-\alpha$.now according to $ \cos$ law in $ \triangle ABC$ and $ \triangle ACD$ we have: $ a^2+d^2-2ad\cos\alpha=AC^2=b^2+c^2+2bc\cos\alpha$ hence: $ 2\cos\alpha=\frac{a^2+d^2-b^2-c^2}{ad+bc}$ \begin{eqnarray*}\Rightarrow AC^2=a^2+d^2-ad.\frac{a^2+d^2-b^2-c^2}{ad+bc}=\frac{(ab+cd)(ac+bd)}{ad+bc}\end{eqnarray*} now note that $ ABCD$ is cyclic,hence according to ptolemy's theorem we get that: $ (AC.BD)^2=(ab+cd)^2$ hence: $ (ac+bd)(a^2-ac+c^2)=(ab+cd)(ad+bc)$ (**) now note that: $ ab+cd>ac+bd>ad+bc$ (***) (the first one follows from $ (a-d)(b-c)>0$ and the second one from $ (a-b)(c-d)$) now assume that $ ab+cd$ is a prime number,first of all from (***) we get that $ ab+cd$ and $ ac+bd$ are relatively prime.so by (**) $ ac+bd$ must devide $ ad+bc$ which is a contradiction according to (***)...

Here is my solution. Let $ P(x)=(x+b)(x+d)-(x-a)(x+c)$ $ \textcolor{white}{Let P(x)}=(b+d+a-c)x+ac+bd$ $ \textcolor{white}{Let P(x)}=(b+d+a-c)(x+b+d-a+c)$. $ P(a)=(a+b)(a+d)=(b+d+a-c)(b+c+d)$. $ P(a)\neq 0$, so $ b+d+a-c \mid (a+b)(a+d)$ but $ a+b>b+d+a-c>\dfrac{a+b}{2}$, therefore a part of $ b+d+a-c$ must divide $ a+d$. Thus, we have $ K=(b+d+a-c,a+d)=(b-c,a+d)=(b-c,b+d+a-c)\neq 1$. Since $ K\mid a+d$, we have $ K\le a+d<ab+cd$. $ (b-c)(a-d)=ab+cd-ac-bd$. Using $ ac+bd=(b+d+a-c)(b+d-a+c)$ we have $ ab+cd=(b-c)(a-d)+(b+d+a-c)(b+d-a+c)$. So $ K\mid ab+cd$ with $ 1<K<ab+cd$, which implies that $ ab+cd$ is composite.

This is similar to BaBaK Ghalebi's solution, but it differs a bit at the end:

EDIT: Sorry, this solution is incorrect. $ BD$ can be in the form of $ \frac {q^2}{p}$ for some $ p$, $ q$ primes. EDIT 2: I hope this works:

I think it is an easy problem for imo6 I think it is good for imo1

ufuk wrote: I think it is an easy problem for imo6 I think it is good for imo1 agreed. The BaBak's solution is quite predictable, upon expanding, the next step gets pretty clear, though that's too tough to be given as imo1.

orl wrote: 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. $ ac + bd = (b + d)^2 - (a - c)^2 \iff$ $ a^2 - ac + c^2 = b^2 + bd + d^2$ Let $ \omega = e^{i\frac {2\pi}{3}}$. Hence $ \omega^3 = 1, \omega^2 + \omega + 1 = 0$. Let's work in the eisenstein integers, ie. $ \mathbb{Z}[\omega]$. $ (a + \omega c)(a + \omega^2 c) = (b - \omega d)(b - \omega^2 d)$ $ (a + \omega c)\overline{(a + \omega c)} = (b - \omega d)\overline{(b - \omega d)} \iff$ $ x = a + \omega c, y = c + \omega a, z = b - \omega d, t = d - \omega b$ Let $ x = a + \omega c, \overline{x} = a + \ c + \omega a, y = b - \omega d, \overline{y} = b - \omega^2 d$ Then $ x \overline{x} = y \overline{y}$ Let $ n = \gcd(x,y)$ and let $ m = \frac {x}{n}$, $ n,m \in \mathbb{Z}$. Obviously $ m \mid \overline{y}$ and hence $ \gcd(x,\overline{y}) = m$. Then $ x = nm, y = n\overline{m}$. $ n = p + \omega q, m = r + \omega s, p,q,r,s \in \mathbb{Z}$ $ x = a + \omega c = (p + \omega q)(r + \omega s) = pr - qs + \omega(qr + ps - qs)$ Hence $ a = pr - qs$ and $ c = qr + ps - qs$. $ y = b - \omega d = (p + \omega q)(r + \omega^2 s) = pr + qs - ps + \omega (qr - ps)$ Hence $ b = pr + qs - ps$ and $ d = ps - qr$ Now $ ab + cd = (pr - qs)(pr + qs - ps) + (qr + ps - qs)(ps - qr)$ Expanding $ ab + cd = p^2r^2 + pqrs - p^2rs - pqrs - q^2s^2 + s^2pq + pqrs + p^2s^2 - pqs^2 - q^2r^2 - pqrs + q^2rs$ $ ab + cd = p^2r^2 - p^2rs - q^2s^2 + p^2s^2 - q^2r^2 + q^2rs = (p^2 - q^2)(r^2 + s^2) - (p^2 - q^2)rs$ $ ab + cd = (p^2 - q^2)(r^2 - rs + s^2)$ Assume that $ ab + cd$ is not a prime. Then either $ p^2 - q^2 = \pm 1$ or $ r^2 - rs + s^2 = \pm 1$. If $ p^2 - q^2 = \pm1$ then $ p = 0,q = \pm 1$ or $ p = \pm 1, q = 0$. If $ p = 0$ then $ a + b = 0$. If $ q = 0$ then $ c = d$. So $ p,q\neq 0 \Rightarrow p^2 - q^2 \neq \pm 1$. $ r^2 - rs + s^2 = \frac {1}{2}(r^2 + (r - s)^2 + s^2) < - 1$ so $ r^2 - rs + s^2 = \pm 1 \iff r^2 - rs + s^2 = 1$. This has the solutions $ (r,s) \in \{(0,1);(0, - 1);(1,0);( - 1,0);(1,1);( - 1, - 1)\}$ If $ r = 0$ then $ b + c = 0$. If $ s = 0$ then $ a = b$. If $ r = s$ then $ a = d$. Hence $ r^2 - rs + s^2 \neq \pm 1$. So $ ab + cd$ are a product of to integers with numerical value at least $ 2$. Hence $ ab + cd$ is composite (Therefore not prime).

orl wrote: Let $a >0, b > c > d\geq0$ be integers and suppose that \[ ac + bd = (b+d+a-c)(b+d-a+c). \] Prove that $ab + cd$ is not prime. $ab+cd=\frac{(b^2-c^2)(b^2+bd+d^2)}{ab-cd-bc},$ where $(b^2+bd+d^2)-(b^2-c^2)=c^2+d^2+bd>0,$ $(b^2-c^2)-(ab-cd-bc)=\frac{(b-c)(c-d)(b^2-c^2+d^2+ad+bd)}{a(a+b-c+d)}>0,$ $ab-cd-bc=\frac{(b-c)[a^2+b^2+c^2+d^2+bc+2da+(b-c)(a+d)]}{2a+b-c+2d}>0.$

See also : http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150525

orl wrote: ..... $16(ab+cd)=(3p^2+n^2)(3m-q)(m+q)$ and it's easy to handle it now. From here, how to show that $ab+cd$ is not prime? Sorry, it still isn't really obvious for me.

orl wrote: Solution by Iura: Lemma: If $ab=cd$ then there are $m,n,p,q$ with $a=mn, b=pq, c=mp, d=nq$. Proof is obvious. Now let's change the notations into $a,b,c,d$ to become more comfortable. Opening the brackets we get $a^2+c^2-ac=b^2+bd+d^2$ Multiplying by $4$ we get $(2a-c)^2+3c^2=(2b+d)^2+3d^2$ thus $3(c+d)(c-d)=(2b+d-2a+c)(2b+d+2a+c)$. Now using this lemma we get either $c+d=mn, c-d=pq, 2a+c-2b+d=3mp, 2b+d+2a+c=nq$ or and notice that $16(ab+cd)=(3p^2+n^2)(3m-q)(m+q)$ and it's easy to handle it now. The second case is analogous. I think it should be $2b+d+2a-c$ instead of $2b+d+2a+c$. Am I right? Could you also elaborate how to obtain $16(ab+cd)=(3p^2+n^2)(3m-q)(m+q)$? Thanks.

Yes, obviously it should be $2b+d+2a-c$ instead of $2b+d+2a+c$. orl made a typo. More importantly, you are also right to ask yourself why from $16(ab+cd) = (3p^2+n^2)(3m-q)(m+q)$ follows that $ab+cd$ cannot be a prime. It does not. For $m=q=2$, that writes $ab+cd = 3p^2+n^2$, and there are infinitely many primes of this form, for example $61 = 3\cdot 2^2 + 7^2$.

One can take $m, q$ in the very beginning such that $(m,q)=1$, so I quite think that the solution proposed is correct, but not entirely clearly formulated. I like the geometric approach. It shows the power of geometry even outside its boundary of usage

orl wrote: Solution by Iura: Lemma. If $ab=cd$ then there exist $m,n,p,q$ with $a=mn, b=pq, c=mp, d=nq$. Proof is obvious. Opening the brackets we get $a^2+c^2-ac=b^2+bd+d^2$ Multiplying by $4$ we get $(2a-c)^2+3c^2=(2b+d)^2+3d^2$ thus $3(c+d)(c-d)=(2b+d-2a+c)(2b+d+2a+c)$. Now using this Lemma we get either \[c+d=mn, c-d=pq, 2a+c-2b+d=3mp, 2b+d+2a+c=nq,\] or \[c+d=mn, c-d=pq, 2a+c-2b+d=mp, 2b+d+2a+c=3nq.\] For the first case, notice that $16(ab+cd)=(3p^2+n^2)(3m-q)(m+q)$, and it's easy to handle it now. The second case is analogous. How can $ab=cd$ given that $a>b>c>d$? And $a,b,c,d>0$ What am I missing.

== Solution == Equality is equivalent to $ a^2 - ac + c^2 = b^2 + bd + d^2 (1) $. Let $ABCD$ be the quadrilateral with $AB = a$, $BC = d$, $CD = b$, $AD = c$, $ \angle BAD = 60^\circ$, and $ \angle BCD = 120^\circ$. Such a quadrilateral exists by $(1)$ and the Law of Cosines. By Strong Form of Ptolemy's Theorem, we find that; $BD^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}$ and by rearrangement inequality; $ab+cd > ac+bd > ad+bc$. Assume $ab+cd = p$ is a prime, since $a^2 - ac + c^2 = BD^2$ is an integer $p \times \frac{ad+bc}{ac+bd}$ must be an integer but this is false since $(p,ac+bd) = 1$ and $ac+bd > ad+bc$. Thus $ab+cd$ can not be a prime.

The solutions that used Four number Lemma posted here are all incomplete(or incorrect). When u get $3(ab+cd)=F\times G$, how could u state that it is a prime?

We can rewrite the equation as \[ac+bd=(b+d)^2-(a-c)^2\implies a^2-ac+c^2=b^2+bd+d^2.\]Interpret this as rewrite wrote: We have a cyclic quadrilateral, two of whose opposite angles are $60$ and $120$. The sides adjacent to the $60$ are $a$ and $c$, and the sides adjacent to the $120$ are $b$ and $d$. Prove that the product of the diagonals of this cyclic quadrilateral must be composite, given $a>b>c>d$ and all four are integers. Let $e$ and $f$ be the length of those diagonals. (Doesn't quite matter what we denote them by.) \[e^2=\frac{(ab+cd)(ad+bc)}{ac+bd}\text{ and } f^2=\frac{(ab+cd)(ac+bd)}{ad+bc}.\]If $e\ne f$, then say $e^2=1$ and $f^2=p^2$. Notice that $e^2=1$ is impossible as $(ab+cd)(ad+bc)>ac+bd$ and similarly if $f^2=1$. Therefore, we must have them equal and their value $p$. But it's impossible for a square to be a prime, since it's an integer by assumption. So we are done. $\blacksquare$

Note that $b+d+a-c \mid ac+bd + a(b+d+a-c) = (a+b)(a+d)$ and similarly $b+d-a+c \mid (c+b)(c+d)$. Let $(a+b)(a+d) = x(b+d+a-c)$ and $(c+b)(c+d) = y(b+d-a+c)$ for integers $x$ and $y$. Multiplying the two equations yields $$(a+b)(c+d)(a+d)(c+b) = xy(ac+bd)$$$$(ac+bd+ad+bc)(ac+bd+ab+cd) = xy(ac+bd)$$Taking modulo $ac+bd$ both sides yields $$ac+bd \mid (ad+bc)(ab+cd)$$Since $a>b>c>d$ we have $ab+cd > ac+bd > ad+bc$ so if $ab+cd$ is prime then $ac+bd \mid ad+bc$, contradiction. Thus $ab+cd$ is not prime.

CLAIM: We have \[ ac + bd | (ab+cd)(ad+bc). \]PROOF: b+d+a-c | (ac+bd) + a(b+d+a-c) = (a+d)(a+b) b+d+a-c | (ac+bd) + c(b+d-a+c) = (c+b)(c+b) So, (b+d+a-c)(b+d-a+c) | (a+d)(a+b)(c+b)(c+b). ac+bd | ((ac+bd) + (ad+bc))((ac+bd)+(ab+dc)) Proved.... NOW, as a>b>c>d, ad+cd > ac+bd > ad+bc (By rearrangement inequality) If ab+cd is prime , then ac+bd | ad+bc which implies ac+bd <ad+bc(or equal to) a contradiction.... Hence, ab+cd is composite.

Our condition can be expanded to be \[a^2-ac+c^2 = b^2+bd+d^2,\] so we can construct cyclic quadrilateral $PQRS$ with $\angle P = 60$, $\angle R = 120$, $PQ=a$, $QR=b$, $RS=d$, and $SP=c$. Hence Strong Ptolemy tells us \[a^2-ac+c^2 = \frac{(ad+bc)(ab+cd)}{ac+bd} \implies (ac+bd) \mid (ad+bc)(ab+cd).\] Our bounding condition says $ad+bc < ac+bd < ab+cd$, so if $ab+cd$ was prime, we get $ac+bd \mid ad+bc$, contradiction. $\blacksquare$

Note $$b+d+a-c\mid ac+bd+a(b+d+a-c)=(a+d)(a+b)$$$$b+d-a+c\mid ac+bd+c(b+d-a+c)=(c+b)(c+d)$$so $$ac+bd\mid \big[(a+d)(c+b)\big]\big[(a+b)(c+d)\big]\Rightarrow ac+bd\mid (ab+cd)(ad+bc)$$We can obtain $ab+cd>ac+bd>ad+bc$ by Rearrangement inequality; so if $ab+cd$ is prime then $ac+bd\mid ad+bc\Rightarrow ad+bc\geq ac+bd$, a contradiction

The expansion is equivalent to $a^2-ac+c^2 = b^2+bd+d^2$. Equivalently, $a, c, d, b$ are the side lengths of a cyclic quadrilateral with opposite angles $60^\circ$ and $120^\circ$. This implies that $b^2+bd+d^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}$ is an integer. Assume for the sake of contradiction that $ab+cd$ is prime; then as $ac+bd > ad+bc$, follows that $ab+cd \mid ac+bd$, but $ac+bd < ab+cd$, contradiction.

Note that $$ac + bd = (b+d-a+c)(b+d+a-c) \iff (a+b)(a+d) = (b+d+c)(b+d+a-c)$$So let's use the Four Number Lemma to conclude: $$\begin{array}{c} a+b =pq \\ a+d =rs \\ b+d+c = ps \\ b+d+a-c =qr \end{array}$$from where we get that $$\begin{array}{c} a=\frac{1}{3}(2pq+2rs-ps-qr) \\ \\ b =\frac{1}{3}(ps+qr+pq-2rs) \\ \\ c=\frac{1}{3}(ps-2qr+rs+pq) \\ \\ d=\frac{1}{3}(ps+qr+rs-2pq) \end{array}$$So we have $$ab+cd = \frac{(2pq+2rs-ps-qr)(ps+qr+pq-2rs)+(ps-2qr+rs+pq)(ps+qr+rs-2pq)}{9} = \frac{r(2p-r)(q^2-qs+s^2)}{3}$$and from $a>b>c>d$ we can easily deduce $p>r$ and $q>s$ so $2p-r\ge 3$ and $q^2-qs+s^2=(q-s)^2+qs \ge 3$. Lastly, we can easily check that if the equality holds for either of them then $r\neq 1$.

The given condition is equivalent to \[a^2-ab+b^2=c^2+cd+d^2 = k,\] which prompt us to construct quadrilateral $PQRS$ with side lengths $a,b,c,d$, such that $PR = \sqrt{k}$. By Law of Cosines, we see that $\angle Q = 60^\circ$ and $\angle S = 120^\circ$, upon which we see that $PQRS$ is cyclic. A stronger form of Ptolemy shows that \[\frac{(ad+bc)(ab+cd)}{ac+bd} \in \mathbb{Z} \implies (ac+bd) \mid (ad+bc)(ab+cd).\] If $ab+cd$ were prime, then $ac+bd \mid ad+bc$, but $ac+bd>ad+bc$, a contradiction. $\square$

We are given that $ac + bd = (b+d-a+c)(b+d+a-c).$ It follows that: $$b+d+a-c|(ac+bd)+c(b+d−a+c)$$The right hand side of this expression expands to: $$ab+ad+a^2+bd$$Which can be factored into: $$(a+b)(c+d).$$Similarly: $$b+d-a+c|(ac+bd)+c(b+d−a+c)$$The right hand side, similar to the first part, expands to: $$cb+cd+c^2+bd.$$This can be factored into: $$(c+b)(c+d).$$Therefore, we have: $$(b+d+c-a)(b+d-c+a)|(a+b)(a+d)(c+d)(c+b).$$We can manipulate the RHS into $((ac+bd)+(ad+bc))((ac+bd)+(ab+dc))$. This yields the expression: $$ac+bd|((ac+bd)+(ad+bc))((ac+bd)+(ab+dc)).$$Therefore: $$ac + bd | (ab + cd)(ad + bc).$$If $ab+cd$ was a prime, it would be coprime to $ac+bd$. However, we have found that is not the case. Therefore, by contradiction $\boxed{ab+cd \text{ is not prime}}$ (I'm still new to writeups so if this has errors please let me know)

MathPanda1 wrote: The condition is equivalent to $a^2-ac+c^2=b^2+bd+d^2$. Thus, there exists a quadrilateral $ABCD$ with $AB=a,BC=c,CD=b,AD=d$ and $ \angle ABC=60^\circ$ and $ \angle CDA=120^\circ$ by Cosine Law. By Ptolemy's Theorem, $ab+cd=AC \cdot BD$. If $AC=1$, then by triangle inequality, $a \leq c+1$, a contradiction. If $BD=1$, then by triangle inequality, $a \leq d+1$, a contradiction. This is because $a>b>c>d$ and $a, b, c, d$ are positive integers. Hence, $AC,BD \geq 2$, so $ab+cd=AC \cdot BD$ is composite, as required. how can you tell AC,BD are integers

Claim: $(ac+bd)(a^2-ac+c^2) = (ab+cd)(ad+bc)$ Consider a quadrilateral $ABC$ with $AB=a, BC=d, CD=b, CA=c$ with $\angle BAD = 60^\circ, \angle BCD = 120^\circ$. Then, the given condition implies: $BD^2 = b^2+bd+d^2=a^2-ac+c^2.$ Note that $ABCD$ is a cyclic quadrilateral. Let $\theta = \angle ABC$. Then: $$a^2+d^2-2ad \cos \theta = b^2+c^2+2bc \cos \theta$$$$\implies \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} = \cos \theta$$By Cosine-Rule: $$AC^2 - a^2+d^2-2ad \cos \theta = \frac{(ac+bd)(ab+cd)}{ad+bc}.$$Notice that, by Ptolemy's theorem: $$AC^2 BD^2 = (ab+cd)^2 $$$$\implies \frac{(ac+bd)(ab+cd)}{ad+bc} \cdot (a^2-ac+c^2)= (ab+cd)^2 $$$$\implies (ac+bd)(a^2-ac+c^2) = (ab+cd)(ad+bc).$$ FTSOC, assume that $ab+cd$ is a prime. Then: $(ac+bd)|(ab+cd)(ad+bc) \implies (ac+bd)|(ad+bc)$. Notice that: $a > b > c > d$. By Re-arrangement inequality: $$ab+cd > ac+bc > ad+bc$$which contradicts that $(ac+bd)|(ad+bc)$. Therefore we arrive at contradiction. Hence $ab+cd$ is not a prime

This great problem with a geometric interpretation was proposed by Alexander Ivanov, Bulgaria.