the only solution is (2,2,3). If c=2, RHS=7 but LHS > 11. Contradiction. If a=2, C^2 + 3 = 4+2b +b^2= (b+1)^2 + 3 So b=c-1. The only two consecutive primes are 2,3. Assume all three a,b,c are odd. LHS is odd, RHS is even, contradiction.

parmenides51 wrote: Determine all three primes $(a, b, c)$ that satisfied the equality $a^2+ab+b^2=c^2+3$. $c=2 \implies a^2+ab+b^2=7$ which has no solution in prime numbers. So, $c$ must be a prime number $>2$, and hence in particular $c$ must be odd, and so $c^2+3$ is even, and so $a^2+ab+b^2$ must be even. Which means both $a,b$ must be even (otherwise $a$ is odd and $b$ is even $\implies a^2$ is odd, $ab$ even and $b^2$ even which means $a^2+ab+b^2$ odd, and if $a,b$ both odd, then $a^2,ab,b^2$ are all odd, and hence their sum is odd). Now, $\implies$ $a,b$ are even prime numbers. So, the only possibility is $a=2,b=2$, and for that we get $a^2+ab+b^2=4+4+4=12=3^2+3$, and hence $(a,b,c)=(2,2,3)$ is the only solution of the equation $a^2+ab+b^2=c^2+3$ in prime numbers.