a) $a_1+a_2+a_3=3n^2-31n+100\equiv n^2+n=n(n+1)\equiv0\pmod2$ b) $a_1$ don't have the same parity as $n$ but $a_3$ have the same parity as $n$ so $a_1=2$ or $a_3=2$ Case 1 : $a_1=2$ so $n^2-10n+21=0$ so $(n-5)^2=4$ so $n=3.7$ we see that there are both solution Case 2 : $a_3=2$ so $n^2-12n+44=0$ so $(n-6)^2=-8$ impossible.

a) $a_1+a_2+a_3=3n^2-31n+100$. If n is even, then $3n^2-31n+100\equiv0\mod2$, or else $3n^2-31n+100\equiv -28n+100\equiv0\mod2$ b)Because sum of three primes are even,one of them must be two i)$n^2-10n+23=2\Rightarrow n=3\hspace{2mm}n=7$. For $n=3$ we got $a_2=13$ and $a_3=19$. For $n=7$ we got $a_2=17$ and $a_3=11$ ii)$n^2-9n+31=2$ No solution. iii)$n^2-12n+46=2$ No solution

for part a : $a_1+a_2+a_3\\=3n^2-31n+100\\ \equiv n^2-n (mod 2)\\ \equiv n(n-1)\\ \equiv 0 (mod 2)$ for part b: lemma: n is odd proof: if n is odd then$ a_3 \equiv 0 (mod 2)$ ==> $n^2-12n+46=2$ this equation has no real solution so our lemma is true Now we have $n \equiv 1(mod 2)$ $a_1=n^2-10n+23 \equiv 1-0+1 \equiv 0(mod 2)$ ==> n^2-10n+23=2 <==>n^2-10n+21=0 <==>n=3,7 putting the values of n in $a_1,a_2,a_3$ respectively we get: $a_1=2 $ $a_2=13,17 $ $a_3=19,11$ hence the required values of n are 3 and 7