user132830: Having perused the stackexchange, I found some similar questions, but am having difficulty understanding how to arrive at the solution to  (n-1)*x=1 mod np, where:
n: Finite group order of the Bitcoin secp256k1 curve
n=115792089237316195423570985008687907852837564279074904382605163141518161494337
p: Prime order of the curve
p=115792089237316195423570985008687907853269984665640564039457584007908834671663
np: (n-1)+(p-1)
np=231584178474632390847141970017375815706107548944715468422062747149426996165998
and (n-1) is not coprime to modulo np.
Having performed the following step of np/2 and adding .5 to result one, so as to achieve:
F1=115792089237316195423570985008687907853053774472357734211031373574713498083000
Then subtracting the initial result with .5 to achieve:
F2=115792089237316195423570985008687907853053774472357734211031373574713498082999
And following instructions from answers to related posts, (n-1) is to be multiplicativeley inversed over mod F1 and F2.  However, neither F1 or F2 are coprime to (n-1). In order to overcome this, it is explained that GCD and CRT are to be used in order to accurately calculate the modular inverse.
What steps are required and how are the operations performed to accomplish this?
Thank you.

user208: What you're asking for is impossible.
You're looking for x such that (n-1)x = 1 mod np. However, both n-1 and np are even. It's easy to see that any multiple of n-1 will be even. Thus, any multiple of n-1 modulo np will also be even. However, the right hand side 1 is odd. This is a contradiction, as the left hand side cannot be odd.

132830: <p>Having perused the stackexchange, I found some similar questions, but am having difficulty understanding how to arrive at the solution to  (<em>n</em>-1)*<em>x</em>=1 <em>mod np</em>, where:</p>
<p><em>n:</em> Finite group order of the Bitcoin secp256k1 curve</p>
<p><em>n</em>=115792089237316195423570985008687907852837564279074904382605163141518161494337</p>
<p><em>p:</em> Prime order of the curve</p>
<p><em>p</em>=115792089237316195423570985008687907853269984665640564039457584007908834671663</p>
<p><em>np:</em> (<em>n</em>-1)+(<em>p</em>-1)</p>
<p><em>np</em>=231584178474632390847141970017375815706107548944715468422062747149426996165998</p>
<p>and (<em>n</em>-1) is not coprime to modulo <em>np</em>.</p>
<p>Having performed the following step of <em>np</em>/2 and adding .5 to result one, so as to achieve:</p>
<p><em>F1</em>=115792089237316195423570985008687907853053774472357734211031373574713498083000</p>
<p>Then subtracting the initial result with .5 to achieve:</p>
<p><em>F2</em>=115792089237316195423570985008687907853053774472357734211031373574713498082999</p>
<p>And following instructions from answers to related posts, (<em>n</em>-1) is to be multiplicativeley inversed over <em>mod F1</em> and <em>F2</em>.  However, neither <em>F1</em> or <em>F2</em> are coprime to (<em>n</em>-1). In order to overcome this, it is explained that GCD and CRT are to be used in order to accurately calculate the modular inverse.</p>
<p>What steps are required and how are the operations performed to accomplish this?</p>
<p>Thank you.</p>
 208: <p>What you're asking for is impossible.</p>
<p>You're looking for <em>x</em> such that <em>(n-1)x = 1 mod np</em>. However, both <em>n-1</em> and <em>np</em> are even. It's easy to see that any multiple of <em>n-1</em> will be even. Thus, any multiple of <em>n-1</em> modulo <em>np</em> will also be even. However, the right hand side <em>1</em> is odd. This is a contradiction, as the left hand side cannot be odd.</p>
