MHB Can Trigonometry and Geometry Prove that ab+pq is Not Prime?

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Prime
AI Thread Summary
The discussion focuses on proving that the expression ab + pq is not prime, given the conditions on integers a, b, p, and q. It presents the equation ap + bq as a product of two factors, highlighting the relationship between these variables. The participants emphasize the application of trigonometric and geometric concepts to solve this number theory problem. The solution showcases the intersection of different mathematical disciplines in addressing prime number properties. Overall, the thread illustrates a creative approach to proving non-primality through geometric and trigonometric reasoning.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Let $a,\,b,\,p,\,q$ be integers with $a>b>p>q>0$.

Suppose that $ap+bq=(b+q+a-p)(b+q-a+p)$.

Prove that $ab+pq$ is not prime.
 
Mathematics news on Phys.org
anemone said:
Let $a,\,b,\,p,\,q$ be integers with $a>b>p>q>0--(1)$.

Suppose that $ap+bq=(b+q+a-p)(b+q-a+p)---(2)$.

Prvoe that $ab+pq$ is not prime.
from (1)(2)we must have:
$p\geq q+1$
$b\geq q+2$
$a\geq q+3$
$ap+bq\geq (q+2+q+q+3-q-1)(q+2+q-q-3+q+1)=(2q+4)(2q)$
is not prime
$\therefore ab+pq\geq (q+2)(q+3)+(q+1)q=2q^2+6q+6$
is not prime
 
Thanks, Albert for participating and also the solution. :)

I want to share with MHB a solution that I thought is great to showcase how trigonometric and geometry skills could be used to tackle a number theory problem such as this one, and I hope you will like it as much as I do:
View attachment 3889

The equality $ap+bq=(b+q+a-p)(b+q-a+p)$ is equivalent to $a^2-ap+p^2=b^2+bq+q^2\tag{1}$.

Let $ABPQ$ be the quadrilateral with $AB=a,\,BP=q,\,PQ=b,\,AQ=p,\,\angle BAQ=60^{\circ}$ and $\angle BPQ=120^{\circ}$.

Such a quadrilateral exists in view of the equation we obtained in (1) and the Law of Cosines.

The common value in (1) is $BQ^2$. Let $\angle ABP=\theta$ so that $\angle PQA=180^{\circ}-\theta$.

Applying the Law of Cosines to triangles $ABP$ and $APQ$ gives

$a^2+q^2-2aq\cos \theta=b^2+p^2+2bp\cos \theta$

Hence $2\cos \theta=\dfrac{a^2+q^2-b^2-p^2}{aq+bp}$ and $AP^2=a^2+q^2-aq\left(\dfrac{a^2+q^2-b^2-p^2}{aq+bp}\right)=\dfrac{(ab+pq)(ap+bq)}{aq+bp}$.

Because $ABPQ$ is cyclic, Ptolemny's Theorem gives $(AP\cdot BQ)^2=(ab+pq)^2$.

It follows that

$\dfrac{(ab+pq)(ap+bq)}{aq+bp}\cdot (a^2-ap+p^2)=(ab+pq)^2$.

or simply $(ap+bq)(a^2-ap+p^2)=(ab+pq)(aq+bp)$.

Next, observe that
$ab+pq>ap+bq>aq+bp\tag{2}$.

The first inequality follows from $(a-q)(b-p)>0$, and the second from $(a-b)(p-q)>0$.

Now, assume that $ab+pq$ is prime, it then follows from (2) that $ab+pq$ and $ap+bq$ are relatively prime. Hence, from $(ap+bq)(a^2-ap+p^2)=(ab+pq)(aq+bp)$, it must be true that $ap+bq$ divides $aq+bp$.

However, this is impossible by (2). Thus $ab+pq$ must not be prime.

One example that satisfies the given conditions is $(65,\,50,\,34,\,11)$.
 

Attachments

  • Prove ab+bq isn't composite.JPG
    Prove ab+bq isn't composite.JPG
    13.3 KB · Views: 91
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top