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Any number not divisible by 2 and not divisible by 3 is in the form: $6k\pm1, \quad k \in \mathbb N$

If I have the following:

$$ p = 6a+b, \qquad a \in \mathbb N, \quad b \in \{-1,1\} $$ $$ q = 6c+d, \qquad c \in \mathbb N, \quad d \in \{-1,1\} $$ $$ n = p \cdot q $$ $$ x = \frac{p+q}{2} $$ $$ y = \frac{q-p}{2} $$ Then: $$ \left(\frac{n+1}{2}\right)^2-x^2=\left(\frac{n-1}{2}\right)^2-y^2=z \qquad \land \qquad \left(\frac{n+1}{2}\right)^2 - \left(\frac{n-1}{2}\right)^2 = n $$ Moreover, for any number $n$ in the form $6k\pm1$, $z$ seems to be divisible by 144. I want to prove that $z$ is divisible by 144.

I will try to probe only one case because there are too many cases.

Supose: $$ \left(\frac{n+1}{2}\right)^2-\left(\frac{p+q}{2}\right)^2 \to \left(\frac{p \cdot q + 1}{2}\right)^2-\left(\frac{p+q}{2}\right)^2 $$ Replacing $p$ and $q$: $$ \left(\frac{\left(6a+b\right)\left(6c+d\right)+1}{2}\right)^{2}-\left(\frac{\left(6a+b\right)+\left(6c+d\right)}{2}\right)^{2} = \left(\frac{36ac+6ad+6bc+bd+1}{2}\right)^{2}-\left(\frac{6a+6c+b+d}{2}\right)^{2} $$ Assuming that $b$ and $d$ are opposite and: $$ b = 1, \quad d = -1 \to \left(\frac{36ac-6a+6c}{2}\right)^{2}-\left(\frac{6a+6c}{2}\right)^{2} $$ Operating and taking common factors: $$ \left(3\left(6ac+c-a\right)\right)^{2}-\left(3\left(a+c\right)\right)^{2} = 9\left(\left(6ac+c-a\right)^{2}-\left(a+c\right)^{2}\right) = $$ $$ = 9\left(36a^{2}c^{2}+12\left(c-a\right)ac+\left(c-a\right)^{2}-\left(a+c\right)^{2}\right) = 9\left(36a^{2}c^{2}+12ac^{2}-12a^{2}c-4ac\right) = $$ $$ 9\cdot4\cdot ac\left(9ac+3c-3a-1\right) $$ If $a$ is even and $c$ is odd: $$ a = 2i, \quad c = 2j - 1 \to 9\cdot4\cdot\left(2i\right)\left(2j-1\right)\left(9\left(2i\right)\left(2j-1\right)+3\left(2j-1\right)-3\left(2i\right)-1\right) = $$

Operating and taking common factors: $$ = 9\cdot4\cdot\left(4ij-2i\right)\left(\left(18i\right)\left(2j-1\right)+\left(6j-3\right)-\left(6i\right)-1\right) = $$

$$ = 9\cdot4\cdot\left(4ij-2i\right)\left(36ij-18i+6j-6i-4\right) = $$

$$ = 9\cdot4\cdot2\cdot2\left(2ij-i\right)\left(18ij-9i+3j-3i-2\right) \to 9\cdot4\cdot2\cdot2 = 144 $$

Finally: $z$ is divisible by 144 and $144 = 12^2$

These means solving these equations: $$ x^2 \equiv \left(\frac{n+1}{2}\right)^2 (\textrm{mod}\ 144) \quad \land \quad y^2 \equiv \left(\frac{n-1}{2}\right)^2 (\textrm{mod}\ 144) $$

You can factorize integer number. But for large integer numbers these method don't work.

So:

Taking any assumptions on $a$, $b$, $c$ or $d$ values or any other assumption always gives that $z$ is divisible by 144. I can't develop a demonstration on all cases. With this I didn't prove anything. I can't refute it either.

Is there an easy way to probe this?

And if this is true, is possible to extend it to factor larger numbers?

Bill Dubuque
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Pedro Urday
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2 Answers2

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Here's an easier way, that proves the general case. Using factoring of the difference of squares and Simon's Favorite Factoring Trick gives

$$\begin{equation}\begin{aligned} z & = \left(\frac{n+1}{2}\right)^2-x^2 \\ & = \left(\frac{pq+1}{2}\right)^2-\left(\frac{p+q}{2}\right)^2 \\ & = \left(\frac{pq+1}{2}+\frac{p+q}{2}\right)\left(\frac{pq+1}{2}-\frac{p+q}{2}\right) \\ & = \frac{(pq + p + q + 1)(pq - p - q + 1)}{4} \\ & = \frac{(p + 1)(q + 1)(p - 1)(q - 1)}{4} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Proving $144 \mid z$ means showing that

$$144(4) \mid (p - 1)(p + 1)(q - 1)(q + 1) \tag{2}\label{eq2A}$$

Note $144(4) = 3^2(2^6) = (3(2^3))^2$. Regardless of $b$ being either $-1$ or $1$, one of $p - 1$ and $p + 1$ is $6a$, so we have $3 \mid (p - 1)(p + 1)$. Also, since $p$ is odd, then $p - 1$ and $p + 1$ are consecutive even integers, so one is congruent to $2 \pmod 4$ while the other is congruent to $0 \pmod{4}$, which means their product has a factor of $2(4) = 8$. Thus, $2^3 \mid (p - 1)(p + 1)$. Similarly, $3 \mid (q - 1)(q + 1)$ and $2^3 \mid (q - 1)(q + 1)$. Combined, this proves that \eqref{eq2A} is always true.

Regarding extending this to factoring larger numbers, if I correctly understand what you mean, then that can also generally be done with appropriate values and expressions.

John Omielan
  • 47,976
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Yes, in fact very easy: applying the well-known $\rm\color{#c00}{C}$ = composition law for differences of squares

$$\ \ \ \color{#0a0}{4z} = (pq\!+\!1)^2-(p\!+\!q)^2\overset{\rm\color{#c00}C} = (p^2\!-\!1)(q^2\!-\!1)\ \ \ \ \ \ $$

So $\,\underbrace{24\cdot 6}_{144}\mid z\iff 24^2\mid \color{#0a0}{4z}=(p^2\!-\!1)(q^2\!-\!1),\,$ true by $\,24\mid n^2\!-\!1\,$ for $n$ coprime to $6$ $\ \ \bf\small QED$

Remark $ $ Just like the e Brahmagupta–Fibonacci identity for composition of sums of squares can be viewed as arising from the multiplicativity of the norms of Gaussian integers, the above composition law fir differences of squares can be viewed as arising from norm multiplicativity of split-complex numbers (aka double or perplex numbers), $\,a+b\:\!j\,$ where $\,j^2 = 1.\ \ $

Bill Dubuque
  • 272,048