6

In reviewing & improving my attempt to solve the question on simultaneous Pell-equations (see here in MSE) I came across a detail which has surely wider implications and which I cannot encompass (correct term?) so far.
The original question is about:
How many solutions (and which) have the simultaneous Pell equations $$ 24a^2+1=t^2 \\ 48a^2+1=u^2 \tag 1 $$

My ansatz, first part: I approached this problem the following way. First look at the difference equation: $$ 24a^2=u^2-t^2 \tag 2 $$ and use the method of parametrization of the pythagorean triples for $b^2 = c^2-a^2$ [which comes out to be $(a,b,c)=(n^2-m^2,2nm, n^2+m^2)$ (wikipedia)] such that I arrive at $$ u=2m^2+3n^2 \\ t=2m^2-3n^2 \\ a= nm \tag 3 $$ Expansion shows that it is indeed a solution for (2)
$$ 24m^2n^2 \overset{?}= (2m^2+3n^2)^2-(2m^2-3n^2)^2 = 12m^2n^2- (-12m^2n^2) = 24m^2n^2 \tag 4 $$ But while this shows solutions for the difference-eq (2) it is easy to see, that this parametric solution for $(a,b,c)$ cannot at the same time be a solution for the first equation $$ 24 a^2+1 = t^2 \tag {5a} $$ which leads to $$ 24(n^2m^2) =(2m^2-3n^2)^2 -1 \tag {5b}$$ This has no solution which can be seen when taken $\pmod 5$.
$\qquad \qquad \qquad $update2: this is corrected from previous version after helpful comment of user servaes, see revision history

So my conclusion on my scribble notes was first: no solutions, simply :-)) ...

But actually there are known solutions for the simultaneous system, namely, $(a,t,u)=(0,\pm1,\pm1)$ and $(\pm1,\pm5,\pm7)$ and thus it looks, as if my approach is useless here and should be discarded.

My ansatz, second part: In a second look on it, after some coffee, I see that this ansatz can be revived, if we allow non-integer values for $(m,n)$. For instance

  • $(m,n)=(\sqrt 3,\frac1{\sqrt 3})$ gives $(a,t,u)=(\frac {\sqrt 3}{\sqrt 3},2\cdot 3-\frac33, 2\cdot 3+ \frac33 )=(1,5,7)$ .
  • Analoguously $(m,n)=(\frac1{\sqrt 2},\sqrt 2,)$ gives $(a,t,u)=(\frac {\sqrt 2}{\sqrt 2},\frac22-3\cdot2, \frac22+ 3 \cdot2)=(1,-5,7)$ .

In this example, the values $\sqrt3$ and $\sqrt 2$ are quite obvious when eq's (3) are considered. But to allow irrational numbers in that well known and such commonly used parametrization of the pythagorean triples shall likely have many more implications- but which I do not see here, even if we uset the restriction that the irrational values have to be selected in a way that $(a,t,u)$ are still all integer.

This is surely of importance, if I want to use the full logic of my small approach for other, but comparable, cases like $Aa^2+1=t^2 ; Ba^2+1=u^2$ .

Q1: can the introduction of that irrational values be understood as meaningful generalization of the Pythagorean-triple parametrization?

Q2: under the assumption of quadratic irrational values for $(n,m)$ - can we still prove, that the set of solutions is that set of the two which are already known?

Gottfried Helms
  • 35,815
  • In the Wikipedia article Pythagorean triple is the quote: "This can be remedied by inserting an additional parameter k to the formula." The parametrization without the $,k,$ is only for primitive triples. Many people overlook this important detail. – Somos Dec 09 '20 at 17:07
  • @Somos - yes, I've seen that; but still the parameter $k$ is integer (well, not explicitely stated, as you make me aware of), and my "Generalization" introduces fractional/irrational values. I fear, this aspect should really be taken into consideration. – Gottfried Helms Dec 09 '20 at 17:37
  • Why is the right-hand side of $(5b)$ not divisible by $3$? It seems to be divisible by $3$ precisely when $m$ is not divisible by $3$. – Servaes Dec 09 '20 at 20:57
  • Also, you parametrization $(3)$ does not cover all solutions. Where did you get it, and why don't you take a complete parametrization? – Servaes Dec 09 '20 at 21:10
  • @Servaes - at first comment: upps, that's a mistake. I think I've transferred that from my scribbles, and didn't check that the parenthese is squared. I'll see what I've done here. At second comment: I do not understand at the moment; do you mean why I didn't mention the third parameter, the common one (the $k$ in wikipedia) at all? – Gottfried Helms Dec 09 '20 at 21:14
  • 1
    A quick scribble suggests that a full parametrization, up to signs and integral multiples, is given by $$(a,t,u)=(mn,6m^2-n^2,6m^2+n^2).$$ In particular it covers your solutions $(a,t,u)=(0,1,1)$ and $(a,t,u)=(1,5,7)$ with $(m,n)=(0,1)$ and $(m,n)=(1,1)$, up to signs. Your parametrization $(3)$ doesn't cover either, so I wonder where and why you got it. This stems from the usual method of parametrization of rational points on a (smooth) conic, and some divisibility considerations to find the integral points. – Servaes Dec 09 '20 at 21:47
  • @Servaes - thanks for that explanation. I see, that the middle term of $(6m^2 \pm n^2)^2$ comes out to be $\pm 12m^2n^2$ which is the same term as in my notation and which suffices to match $24a^2$. Didn't notice it, and thus that this parametrization is not unique. I'll check in which way that two parametrizations differ for the solutions in (1). (Hoping my fiddling has not been complete bs ;-( ) – Gottfried Helms Dec 09 '20 at 21:57
  • Moreover, from this parametrization it quickly follows that if $(a,t,u)$ is a solution, corresponding to $(m,n)$ in the parametrization above, then wlog $$24(mn)^2+1=(6m^2-n^2)^2,$$ which leads to the Thue equation $$36m^4-36m^2n^2+n^2=1,$$ for which standard solution methods (I used PARI/GP) tell me that $(\pm1,\pm1)$ and $(0,\pm1)$ are the only solutions (for all choices of signs, so $6$ total). This means the aforementioned triplets $(a,t,u)$ are the only solutions. – Servaes Dec 09 '20 at 21:58
  • Well, applying your parametrization $(6m^2 \pm n^2)^2$ the insertion of $m=\sqrt 6, n=1/m$ would as well lead to a/to the correct result $(a,t,u)=(1,-5,7)$, without Thue (I've never worked with the Thue equation so far... ). I'll lookout whether my proposal with the irrational parametrizing is possibly nothing else than Thue in disguise. Thanks for that hint. (I use Pari/GP as main software and can thus work & learn with example computations) – Gottfried Helms Dec 09 '20 at 22:16
  • 1
    The point is that Thue's theorem tells us that that quartic equation has only finitely many solution, and gives an effective method to find them all. (And then I use a computer to actually find all six of them). Also, for $m=\sqrt{6}$ and $n=\tfrac1m$ I get $(a,t,u)=(1,\tfrac{1295}{36},\tfrac{1297}{36})$. Edit: I see now that switching $m$ and $n$ does give that solution. – Servaes Dec 09 '20 at 23:19
  • @Servaes - ah yes, the last one was a typo; of course $m=\frac1{\sqrt6} ; n=1/m$ was the correct idea. – Gottfried Helms Dec 10 '20 at 06:51

0 Answers0