MHB Is ab+cd Not Prime in This Integer Puzzle?

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion revolves around a mathematical problem involving integers a, b, c, and d, where a > b > c > d > 0. The equation ac + bd = (b + d + a - c)(b + d - a + c) is presented, leading to the challenge of proving that ab + cd is not a prime number. Participants engage in solving the problem, with Olinguito providing a correct solution. The thread emphasizes the importance of understanding the relationships between the variables to arrive at the conclusion. The mathematical proof demonstrates that under the given conditions, ab + cd cannot be prime.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Here is this week's POTW:

-----

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.

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
Congratulations to Olinguito for his correct solution, which you can find below!(Cool)

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.
 
Back
Top