Using the ring of real quaternions as a model, we define the quaternions over the integers $\text{mod}$ $p,$ $p$ an odd prime number, in exactly the same way; however, now considering all symbols of the form $a_0 + a_1i + a_2j + a_3k,$ where $a_0,a_1,a_2,a_3$ are integers mod $p.$ Prove that $R$ is not a division ring.
I tried solving this problem as follows:
This ring is said to be a division ring, if, the non-zero elements of the ring, form a group under multiplication.
Closure, Idenity and Associative property is satisfied in $R$ as $1+0i+0j+0k$ is the identity element.
We need to show that the inverse of all the non-zero elements doesn't necessarily exist. If possible, let $a^{-1}=b$ exist and be equal to $b=b_0+b_1i+b_2j+b_3k.$
We note that,
$$ab=(a_0+a_1i+a_2j+a_3k)(b_0+b_1i+b_2j+b_3k)=(a_0b_0-a_1b_1-a_2b_2-a_3b_3)\pmod p+i(a_0b_1+a_1b_0+a_2b_3-a_3b_2)\pmod p+j(a_0b_2-a_1b_3+a_2b_0+a_3b_1)\pmod p+k(a_0b_3+a_1b_2-a_2b_1+a_3b_0)\pmod p.$$
Now, as $b$ is the inverse of $a$ so, $$(a_0b_0-a_1b_1-a_2b_2-a_3b_3)\pmod p=1,$$$$(a_0b_1+a_1b_0+a_2b_3-a_3b_2)\pmod p=0,$$$$(a_0b_2-a_1b_3+a_2b_0+a_3b_1)\pmod p=0,$$ $$(a_0b_3+a_1b_2-a_2b_1+a_3b_0)\pmod p=0\tag 1.$$
To prove the asserted claim, we have to show that the system $(1)$ does not always have a solution in $a_i,b_i$.
But this is actually, the point, where I am stuck. I dont understand how to proceed further.