Determine all primes $p>0$ so that there exists a solution in $\mathbb F_p$, the field with $p$ elements. $$a)\ 3x^2 - 30x+40=0$$ $$b)\ x^2 + 26x+139=0$$ $$c)\ x^2 - 5x+6=0$$
For a), I got p=2, p=13. For b), I got p=5, p=31, for c), I got p=3. Could someone please tell me if there is an easier way to calculate all the prime p instead of brute forcing it? Could you also please let me know if these are all the primes p that satisfies the condition?
Thank you!!!
In c) we have $$x^2-5x+6=(x-2)(x-3)$$ so $x=2$ and $x=3$ are solutions modulo any prime $p$.
In b) we have $$ x^2+26x+139=(x+13)^2-30. $$ So this quadratic has a solution modulo $p$, if and only if $30=2\cdot3\cdot5$ is a quadratic residue modulo $p$. In other words we need an even number of $2,3,5$ to be quadratic non-residues modulo $p>5$. By the law of quadratic reciprocity i) the question about $2$ depends on the residue class of $p$ modulo $8$, ii) the question about $3$ depends on the residue class of $p$ modulo $12$, and iii) the question about $5$ depends on the residue class of $p$ modulo $5$. A consequence of this is that the answer depends on the residue class of $p$ modulo $\mathrm{lcm}(8,12,5)=120$. Leaving it to you to check how the $\phi(120)=32$ residue classes split (it works for exactly one half of them and doesn't work for the other half).
Part a) is similar to part b). Leaving that to you as well. Apart from the factors of the discriminant (that you need to check separately), there will be a fifty-fifty split among the primes according to their residue class modulo a suitable number.