I was wondering if any of you experts out there would consider the following to be of any merit.
Given a composite integer, the product of two unique primes (p=nq), an isosceles trapezoid can be constructed where the Area of trapezoid is given as p, the larger base as (p+1)/2, the smaller base as (p-1)/2, with a height of 2. For example, for the trivial composite, p=33, larger base = 17, smaller base=16, height = 2, area = 33.
Such a trapezoid can be manipulated, reducing the size of the two bases while at the same time increasing the height, keeping all angles the same, resulting in a trapezoid with same area, p.
My conjecture (if indeed, it qualifies as a conjecture): There exists only one transformed trapezoid of the same area, p, where the values of the (reduced) bases are both of integer values, and that these values point to the factorization of p, in that base 1 + base 2 = one prime factor (p), and base 1 - base 2 = the other prime factor (q).
In the example of p=33 above, these reduced bases are 7 and 4, respectively, indicating the prime factors of 33 are 11 and 3.
I guess my question is, should I continue to pursue this line of inquiry, or am I simply stating some identity property? I realize this is basically a difference of squares approach. It’s just that I haven’t run across any other “geometric” approach to factoring in my research.
FYI, I have played with this conjecture using geogebra©️ software, constructing and manipulating various “composite” area trapezoids, finding no examples contradicting the conjecture. The limitations of the mentioned software regarding number size, as well as my computer’s clock speed have proven the only obstacles to using this method in factoring large composites.
Thanks in advance for your thoughtful replies.
If we follow your description and let $a$ denote the longest base and $b$ the shortest base and use $h$ to denote the height, then we have: $$ h\cdot{a+b \over 2}=n $$ where $$ b=a-{h \over 2} $$ Thus clearly $h\in 2\mathbb N$ for both $a,b$ to be integers. Combining those equations and solving for $a$ we get $$ a={n\over h}+{h\over4} $$ from which it follows that $$ b={n\over h}-{h\over4} $$ Now since $h/4$ is contained in $\mathbb N/2$ we see that $a,b$ can only be integers if $n/h$ is also contained in $\mathbb N/2$. Hence $h$ divides $2n$. Thus if $n=pq$ we must have $h\in\{2,2p,2q,2n\}$. If $h=2$ we get $$ a=\frac{n+1}{2},b=\frac{n-1}2 $$ If $h=2q$ we get $$ a=\frac{p+q}2,b=\frac{p-q}2 $$ and if $h=2p$ then $p,q$ change roles above. But $b$ will only be positive if $h$ equals twice the smallest prime factor of $n$. Similarly $h=2n$ renders $b$ negative. This proves your conjecture to be correct.