An example in $Z[i√6]$ such that the gcd of two non-zero elements is $1$ but the gcd cannot be expressed as linear combination of the two elements

Question

I need to find an example for the following:

Two non-zero elements $a$ and $b$ in $Z[i\sqrt{6}]$ for which $gcd(a,b)=1$ but there exist no $\alpha$,$\beta$, such that, $a\alpha+b\beta$=1

Now, I think $5$ and $2+i\sqrt{6}$ have $gcd=1$ but I am not able to prove that 1 cannot be expressed as linear combination of the two elements.

I am not sure about the example I have given and even if it is correct I cannot solve the latter part. Any help will be highly helpful as I have exam on this topic on $20^{th}$ and I am still not able to solve this question.

Thanks in advance.

Answer 1

Just do it the hard way.

Take $5(a + b \sqrt{-6}) + (2 + \sqrt{-6})(c + d \sqrt{-6})$ and expand it out. We get $$ (5 a + 2 c - 6 d) + (5 b + c + 2 d)\sqrt{-6}. $$ So in order to get $1$, we want $5a+2c-6d=1$ and $5b+c+2d=0$.

For a slick way to prove that this is impossible, we could take $2\times$ the first equation plus $1\times$ the second equation, and get $$10a + 5b + 5c - 10d = 2$$which has no integer solutions: the left-hand side is divisible by $5$, but the right-hand side is not.

But if we hadn't spotted this, we could just go ahead and solve the two systems of equations to get an infinite family of solutions. Solving for $c,d$ in terms of $a,b$, we get for instance $c = \frac15 - a - 3b$, which obviously can't work for three integers $a,b,c$.

Answer 2

Hint: $\ (1) = (2,\sqrt{-6})\overset{\rm Euclid}\Longrightarrow\,(1)=(2,\sqrt{-6}^2) = (2,-6) = (2)\ \Rightarrow\!\Leftarrow$