Showing a system of equations has no integer solutions

Question

I am having trouble finding an elegant way of showing this system of equations has no integer solutions:

$$32c+11d=9a+10b$$

$$2c+d=a$$

$$ad-bc=1$$

So far I've narrowed it down to showing $221c^2+100$ cannot be a perfect square, but I'm hoping there's a way that's less computational!

Answer 1

ADDED: pretty much everything needed for this answer is in this chapter of BUELL

The method suggested in comments gives first $a = d + 2c$ and then $d = 5b-7c.$ Plugging those into $ad-bc=1$ gives $$ 25 b^2 - 61bc + 35 c^2 = 1. $$ If we had $b=c = 1$ the quadratic form would evaluate to $-1.$ However, $1$ itself is impossible. The printout below shows the Gauss-Lagrange method of "reduced" indefinite binary quadratic forms. It is a theorem of Lagrange that all small numbers (below $\frac{1}{2} \sqrt {221}$ in absolute value) that are primitively integrally represented by $\langle 25 -61, 35 \rangle$ must appear as first or last coefficients of a form in the chain of reduced forms equivalent to the original; however

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle  25 -61 35

  0  form             25         -61          35  delta     -1
  1  form             35          -9          -1  delta     -2
  2  form             -1          13          13


          -1           2
          -1           1

To Return  
           1          -2
           1          -1

0  form   -1 13 13   delta  1     ambiguous  
1  form   13 13 -1   delta  -13     ambiguous  
2  form   -1 13 13


  form   -1 x^2  + 13 x y  13 y^2 

minimum was   1rep   x = 1   y = 0 disc 221 dSqrt 14  M_Ratio  196
Automorph, written on right of Gram matrix:  
-1  13
1  -14
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 

This is equivalent to the fact that $x^2 - 221 y^2 \neq -1$ for integers $x,y,$ proof by continued fractions, really:

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./Pell 221
Tue Jun 30 15:05:16 PDT 2020


0  form   1 28 -25   delta  -1
1  form   -25 22 4   delta  6
2  form   4 26 -13   delta  -2
3  form   -13 26 4   delta  6
4  form   4 22 -25   delta  -1
5  form   -25 28 1   delta  28
6  form   1 28 -25

 disc 884
Automorph, written on right of Gram matrix:  
97  2800
112  3233


 Pell automorph 
1665  24752
112  1665

Pell unit 
1665^2 - 221 * 112^2 = 1 

=========================================

  4 PRIMITIVE 
15^2 - 221 * 1^2 = 4 

=========================================

221      13 *  17

Tue Jun 30 15:05:16 PDT 2020
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 

Answer 2

It is not true because $(a,b,c,d)=(0,-1,-1,2)$ is an integer solutions as it is easily verified.

►$32c+11d=9a+10b$ has a solution $(a,d)=(5b+6c+11n,5b+2c+9n)$ where $n\in\mathbb Z$

►Therefore $2c+d=a$ gives $c=-n$

►It follows $ad-bc=1$ gives $25b^2+59bn+35n^2=1$ which admits the solution $(b,n)=(-1,1),(1,-1)$ from which the above counterexample $(a,b,c,d)=(0,-1,-1,2)$.

Answer 3

By eliminating $a,d$ we get equation $$25 b^2 - 61 b c + 35 c^2=1\tag{1}$$

Solve $(1)$ by modulo $5$ we get $5\not\mid c$, (WA).

$(1)\implies (50 b - 61 c)^2 - 221 c^2 = 100\overset{50 b - 61 c\to X}{\implies}$

$$X^2-221c=100\tag{2}$$

All solutions of Pell equation $(2)$ we get from modulo polynomial $n\cdot u^j$, where $n$ is norm (fundamental solution), $u$ is fundamental unit and $j\in\mathbb{N}$.

For $(2)$ we have $n=10$ and $u\equiv(x-15)/2\pmod{x^2-221}$.

Example:

$10\cdot\Bigl((x-15)/2\Bigr)^{7}\pmod{x^2-221}\equiv 55694245x - 827954475\pmod{x^2-221}$,

i.e. $(X,c)=(827954475,55694245)$.

gp-code: 10*Mod((x-15)/2, x^2-221)^7=Mod(55694245*x - 827954475, x^2 - 221).

Solving $(2)$ in pari/gp:

abcd()=
{
 D= 221; C= 100;
 Q= bnfinit('x^2-D, 1);
 fu= Q.fu[1]; print("Fundamental Unit: "fu);
 N= bnfisintnorm(Q, C); print("Fundamental Solutions (Norm): "N"\n");
 for(i=1, #N, ni= N[i];
  for(j=0, 16,
   s= lift(ni*fu^j);
   X= abs(polcoeff(s, 0)); Y= abs(polcoeff(s, 1));  
   if(Y, if(X^2-D*Y^2==C,
    print("("X", "Y")    j="j)
   ))
  )
 )
};

Output:

? abcd()
Fundamental Unit: Mod(1/2*x - 15/2, x^2 - 221)
Fundamental Solutions (Norm): [10]

(75, 5)    j=1
(1115, 75)    j=2
(16650, 1120)    j=3
(248635, 16725)    j=4
(3712875, 249755)    j=5
(55444490, 3729600)    j=6
(827954475, 55694245)    j=7
(12363872635, 831684075)    j=8
(184630135050, 12419566880)    j=9
(2757088153115, 185461819125)    j=10
(41171692161675, 2769507719995)    j=11
(614818294272010, 41357153980800)    j=12
(9181102721918475, 617587801992005)    j=13
(137101722534505115, 9222459875899275)    j=14
(2047344735295658250, 137719310336497120)    j=15
(30573069306900368635, 2056567195171557525)    j=16

In fine, because for $(2)$ norm $=10$, then $5\mid c$ for $(2)$. But this contradiction with $5\not\mid c$ for $(1)$.