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I would like to find all integers $a$ such that $x^5-x-a$ has a quadratic factor in $\mathbb{Z}[x]$.

My Attempt

Let $x^5-x-a=(x^2+bx+c)(x^3+dx^2+ex+f)$, so that we have the following:

$$\begin{array}{rcl} b+d&=&0\\ e+bd+c&=&0\\ f+be+cd&=&0\\ bf+ce&=&-1\\ cf&=&-a \end{array}$$

Hence:

$$\begin{array}{rcccl} d&=&-b\\ e&=&-bd-c&=&b^2-c\\ f&=&-be-cd&=&-b^3+2bc \end{array}$$

and we have:

$$1=-bf-ce=b^4-3b^2c+c^2,$$

so that:

$$(2c-3b^2)^2=5b^4+4.$$

Question

How can I find all values of $n$ such that $5n^4+4$ is a perfect square?

My Attempt

If $m^2=5n^4+4$, then $m^2-5n^4=4$.

If $m=2m_*$, then $n$ is even, so that $n=2n_*$, and we have the equation $m_*^2-20n_*^4=1$. By Pell equation, since $(a,b)=(9,2)$ is the least non-trivial solution of $a^2-20b^2=1$, then the general solution has the form $(a_n,b_n)$ where $a_n+b_n\sqrt{20}=(9+2\sqrt{20})^n$, but I do not know how to find out what values of $n$ make $b_n$ a square.

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Short version: in $w^2 - 5 v^2 = 4,$ the numbers $v$ are Fibonacci numbers, of which the largest perfect square is $144$

As you can see, my "v" numbers are alternate Fibonacci numbers, while "w" are Lucas. I will try to find a reference, it is known that the largest square Fibonacci number is 144. Your largest $n$ is therefore $12,$ where your $m=322$

COHN 1963

Umm. Here is a Conway topograph for the quadratic form $x^2 - 5 y^2.$ This constitutes a proof that all solutions of $x^2 - 5 y^2 = 4$ are generated by initial pairs $$ (2,0) , (3,1) , ( 7,3), (18,8), (47,21), 123,55), (322,144), (843, 377) $$ with recursions $$ x_{n+6} = 18 x_{n+3} - x_n $$ $$ y_{n+6} = 18 y_{n+3} - y_n $$

These are from Cayley-Hamilton for $$ \left( \begin{array}{cc} 9&20 \\ 4&9 \end{array} \right) $$

A little more work shows that we may interpolate, meaning $$ x_{n+2} = 3 x_{n+1} - x_n $$ $$ y_{n+2} = 3 y_{n+1} - y_n $$

Let's see, the irrationals in the Binet description of alternate Fibonacci numbers are $$ \frac{3 \pm \sqrt 5}{2}, $$ while $$ \left(\frac{3 \pm \sqrt 5}{2} \right)^3 = 9 \pm 4 \sqrt 5 $$ where $9 \pm 4 \sqrt 5$ are the Binet numbers from $\lambda^2 - 18 \lambda + 1 =0$

enter image description here

REsources on Conway's Topograph

http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf (Conway)

https://www.math.cornell.edu/~hatcher/TN/TNbook.pdf (Hatcher)

http://bookstore.ams.org/mbk-105/ (Weissman)

http://www.springer.com/us/book/9780387955872 (Stillwell)

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  Automorphism matrix:  
    9   20
    4   9
  Automorphism backwards:  
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    -4   9

  9^2 - 5 4^2 = 1

 w^2 - 5 v^2 = 4 =  2^2

Thu Nov 19 10:06:08 PST 2020

w:  2  v:  0  SEED   KEEP +- 
w:  3  v:  1  SEED   KEEP +- 
w:  7  v:  3  SEED   BACK ONE STEP  3 ,  -1
w:  18  v:  8  SEED   BACK ONE STEP  2 ,  0
w:  47  v:  21
w:  123  v:  55
w:  322  v:  144
w:  843  v:  377
w:  2207  v:  987
w:  5778  v:  2584
w:  15127  v:  6765
w:  39603  v:  17711
w:  103682  v:  46368
w:  271443  v:  121393
w:  710647  v:  317811
w:  1860498  v:  832040
w:  4870847  v:  2178309
w:  12752043  v:  5702887

Thu Nov 19 10:06:37 PST 2020

 w^2 - 5 v^2 = 4 =  2^2
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  • 2
    $\begingroup$ Surely coincidence that the Fibonacci numbers show up for a degree-5 problem, but quite an elegant coincidence! $\endgroup$ – Steven Stadnicki Nov 19 at 18:30
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This is essentially an elliptic curve.

There might be elementary methods, but there are also computer algebra systems that can (in many cases) solve this kind of diophantine equations.

We may rewrite the equation as: $m^2n^2 = 5n^6 + 4n^2$.

If we write $y = 5mn$ and $x = 5n^2$, then it becomes $y^2 = x^3 + 20x$.

Now we use Sage to find all integer points on this curve. Paste the following codes into this site and press "Evaluate".

EllipticCurve([20, 0]).integral_points()

The output:

[(0 : 0 : 1), (4 : 12 : 1), (5 : 15 : 1), (720 : 19320 : 1)]

We see that the corresponding values of $(m, n)$ are $(2,0), (3,1), (322,12)$, respectively (negative values are not listed).

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  • $\begingroup$ What do you qualify as an elliptic curve? If that is true then there are only a finite number of possible values for the constant $a?$. $\endgroup$ – Piquito Nov 19 at 20:11
  • $\begingroup$ Yes, I have given here all integral solutions to the equation $m^2 = 5n^4 + 4$ (up to signs). An elliptic curve is a (projective, smooth) curve of genus one, together with a rational point. Here we actually have just a curve of genus one, and I chose a (birational) Weierstrass model so that elliptic curve solvers could be used. $\endgroup$ – WhatsUp Nov 20 at 0:19
  • $\begingroup$ Dear friend: Your rational point in your definition is the trivial one (the zero) because a lot of elliptic curves have no (non-trivial) rational points. On the other hand, the genus of $y^2=5x^4+4$ seems to be greater than $1$ (no singularity). I would like to know what your Weirstrass model is. Sincerely regards. $\endgroup$ – Piquito Nov 21 at 9:46
  • $\begingroup$ For generalities on elliptic curves, please refer to the GTM book of J. Silverman. There are too many things to explain, which cannot be all done in a comment. I can only confirm that what I stated here is correct. $\endgroup$ – WhatsUp Nov 21 at 14:43
  • $\begingroup$ You could give just the genus of your curve, which should be equal to $1$ (a birational relationship preserves the genus). Regards.. $\endgroup$ – Piquito Nov 21 at 15:00
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Another way is to do a long division of $x ^ 5-x-a$ by the arbitrary trinomial $x ^ 2 + bx + c$ and set the remainder to zero. This gives the remainder $$(c ^ 2-3b ^ 2c + b ^ 4-1) x + (cb ^ 3-2bc ^ 2-a) = 0$$ from where we have $a = cb (b ^ 2-2c)$ and $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$.

This means that for every solution of $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$ we have a corresponding value $a = cb (b ^ 2-2c)$.

Some solutions of $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$ are $(b,c)=(1,3),(0,1),(12,55),(12,377)$.

EXAMPLES.-$(b,c)=(1,3)$ gives $a=-15$ and we have $$x^5-x+15=(x^2+x+3)(x^3-x^2-2x+5)$$ $(b,c)=(12,377)$ gives $a=-2759640$ and we have $$x^5-x+2759640=(x^2+12x+377)(x^3-12x^2-233x+7320)$$

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