Showing a system of equations has no integer solutions
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
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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
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4 PRIMITIVE
15^2 - 221 * 1^2 = 4
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221 13 * 17
Tue Jun 30 15:05:16 PDT 2020
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$