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

[anemone]

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.

 

[Albert1]

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.

Spoiler

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

 

[anemone]

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:

Spoiler

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 $(65,\,50,\,34,\,11)$.