Values of $a$ such that $x^5-x-a$ has quadratic factor

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 $\lambda^2 - 18 \lambda + 1 =0$

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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jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:  
    9   20
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  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

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).

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 $$x^5-x+2759640=(x^2+12x+377)(x^3-12x^2-233x+7320)$$