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While working on another problem, I stumbled on the following divisibility claim.

Conjecture: No integers $a,b,c,d$ satisfy all of the following conditions:

  1. $a^2+b^2-c^2-d^2 = 2(ad-bc)-1$;
  2. $\gcd(ac+bd,a^2+b^2-c^2-d^2) > 1$;
  3. $\gcd(ac+bd+1,ad-bc) > 1$;
  4. $(ac+bd)-(ad-bc) = \gcd(ac+bd,ad-bc)^2 > 1$.

There is also the following weaker version.

Conjecture (Weak Version): No integers $a,b,c,d$ satisfy all of the following conditions:

  1. $a^2+b^2-c^2-d^2 = 2(ad-bc)-1$;
  2. $\gcd(ac+bd,a^2+b^2-c^2-d^2) > \sqrt{ac+bd}$;
  3. $\gcd(ac+bd+1,ad-bc) > 1$;
  4. $(ac+bd)-(ad-bc) = \gcd(ac+bd,ad-bc)^2 > ad-bc$.

It seems like either (and especially the Weak Version) should be easy to prove… but I've had no luck. Note that all four restrictions seem necessary, as removing any one allows solutions to be found.

Anyone have any hints for me?

EDIT #1:

We have the identity \begin{align*} b^2(a^2+b^2-c^2-d^2) - (ac+bd)(ac-bd) = (a^2+b^2)(b^2-c^2), \end{align*} so $\gcd(a^2+b^2-c^2-d^2\!,\, ac+bd) \mid (a^2+b^2)(b^2-c^2).$ We also have \begin{align*} a(ac+bd)-b(ad-bc) = c(a^2+b^2) \end{align*} so $\gcd(ac+bd,\, ad-bc) \mid c(a^2+b^2)$. Finally, we have \begin{align} a^2\bigl((ac+bd+1)-(a^2+b^2-c^2-d^2+1)\bigr) - (ad-bc)(ad+bc+ab) = (a^2+b^2)(c^2+ac-a^2), \end{align} so $\gcd(ac+bd+1,\, ad-bc)\mid (a^2+b^2)(c^2+ac-a^2)$; alternatively [and more efficiently], \begin{align} a(ac+bd+1)-a(ad-bc) = c(a^2+b^2)+a, \end{align} so $\gcd(ac+bd+1,ad-bc) \mid \bigl(c(a^2+b^2)+a\bigr)$. But I can't seem to take it to the goal line.

EDIT #2:

If we write $g=ac+bd$ and $f=ad-bc$, replace Conditions 2-4, and “absorb” Condition 1, we have

  1. $\gcd(g,2f-1) > 1$;
  2. $\gcd(g+1,f) > 1$;
  3. $g-f = \gcd(g,f)^2 > 1$.

This system has many solutions! That is to say, not only can Condition 1 not be eliminated entirely, it cannot even be “absorbed” into the other conditions.

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  • $\begingroup$ just for curiosity, did you try to put it as gaussian integers / vectors? $\endgroup$ – G Cab Apr 27 '16 at 14:37
  • $\begingroup$ @GCab: No — wanted an totally elementary proof. I believe [hope!?] I've now found it. $\endgroup$ – Kieren MacMillan Apr 27 '16 at 15:51
  • $\begingroup$ @GCab: But any proof would be helpful! $\endgroup$ – Kieren MacMillan Apr 27 '16 at 18:06
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Not an answer, but just to share if it could be a way, on the track of your edit2.
Put $z = a + i\,b\quad w = c + i\,d$ then: $$ a^{\,2} + b^{\,2} - c^{\,2} - d^{\,2} = \tilde z\,z - \tilde w\,w $$ $$ \left( {\tilde z + \,\tilde w} \right)\left( {z - w} \right) = \tilde z\,z - \tilde z\,w + \tilde w\,z - \tilde w\,w = \tilde z\,z - \tilde w\,w - 2{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right) $$ $$ ad - bc = {\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right)\quad ac + bd = {\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right) $$ and $$ \left\{ \matrix{ \tilde z\,z - \tilde w\,w - 2{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right) = \left( {\tilde z + \,\tilde w} \right)\left( {z - w} \right) = - 1 \hfill \cr \gcd \left( {{\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right),\;2{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right) - 1} \right) > 1 \hfill \cr \gcd \left( {1 + {\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right),\;{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right)} \right) > 1 \hfill \cr {\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right) - {\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right) = \gcd \left( {{\mathop{\rm Re}\nolimits} \left( {\tilde z\,w} \right),\;{\mathop{\rm Im}\nolimits} \left( {\tilde z\,w} \right)} \right)^{\,2} > 1 \hfill \cr} \right. $$

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