Prime always make me Confuse

Let the roots of the quadratic equation $x^{2}-63x+c$ be $a$ and $b$ where $a$ and $b$ are prime numbers.

Then, by Vieta's Formula, we have $a+b=63$ and $ab=c$ .

$63$ is an odd number and so $a+b$ has to be an odd number. So, exactly one of them has to be even and one odd because if we add an odd and an even number, we always get an odd number.

Let us consider $a$ to be even. Also, $a$ is a prime number as given in the question. So, $a$ satisfies the condition of an even prime number which is $2$ . Hence, $a=2$ . Putting the value of $a$ in the above equation, we get the value of $b$ as $61$ .

Now, $c=2\times 61=122$ . So, there is only one value of $c$ satisfying the given condition.

Hence,the number of admissible values of $c$ is $\huge{1}$ .

The two roots have to add up to be 63. Since 63 is an odd number, the only possibility is odd + even. Now both have to be prime and how many even numbers are prime? The answer is 1 as 2 is the only even prime. The other root will be 63-2 = 61 which is indeed a prime number too. So this solution is valid and we only have $\boxed{1}$ solution as there is only 1 even prime.