Primes $(a,b,c)$ such that $a^2+b^2+c^2-ab-ac-bc=28$?

Question

This problem was the first version of this one which was simplified by the O.P. to its actual version.

For what prime values of (a,b,c) the expression $f(a,b,c)=a^2+b^2+c^2-ab-ac-bc$ is equal to 28?.

I wonder there are infinitely many solutions. Here my approach.

For $a$ fixed we have an ellipse whose semi-axes measure, $\sqrt{56}$ the largest (located at the diagonal) $y=x$, and $\sqrt{\dfrac{56}{3}}$ the smallest. When $a$ changes the ellipse simply moves along the diagonal $y=x$.

In the attached figure it is shown first the original ellipse (in red) and its normal form (in black) when $a=11$ and second, two ellipses in the original form for the values $a=11$ and $a=23$ (in this second figure the scala has been changed in order to have the figure well set up)

We take as examples of the approximation to the method we want to apply, two ellipses corresponding to the values $a=17,19$ (we already know that the two ellipses have the same size;look at the figure below).

For $a=17$ (red in the figure ) one has the six integer solutions $(a,b,c)=\boxed{(17,23,11)},(17,15,11),\boxed{(17,13,19)},(17,21,15),\boxed{(17,23,19)},(17,21,23)$ (there are more all just changing $(b,c)$ by $(c,b))$ where three of them are prime so solutions of the problem.

For $a=19$ (green in the figure) we have the six integer solutions $(a,b,c)=(19,15,13),\boxed{(19,17,13)},(19,21,15),\boxed{(19,23,17)},(19,25,21),(19,25,23)$ (there are more all just changing $(b,c)$ by $(c,b))$ where two of them are prime so solutions of the problem.

The idea for a possible proof is to apply Dirichlet's theorem of the arithmetical progression because we know that for distinct values of $a$ (always take with prime values), the ellipse is the same, only moving, as above said, along the diagonal $y=x$.

Answer 1

$a^2+b^2+c^2-ab-ac-bc=28$

$a(a-b)+b(b-c)+c(c-a)=28$

$a(a-c)+b(b-a)+c(c-b)=28$

$(a-b)^2+(b-c)^2+(a-c)^2=56$

$x^2+y^2+z^2=56$

$x+y=z$

$x^2+y^2+(x+y)^2=56$

$x^2+y^2<56$

$x,y\in \mathbb N, x^2+y^2<56\implies x^2+y^2 \in \{1,2,4,5,8,9,10,13,16,17,18,20,25,26,29,32,34,36,37,40,41,45,49,50,52,53 \}$

$56=1+55=4+52=9+47=16+40=25+31=36+20=49+7$

$z^2\in \{4,16,36,\}\implies z \in \{\pm2,\pm4,\pm6\}$

$x^2+y^2=52=6^2+4^2+2^2\implies x=6, y=-4, z=2$

$x^2+y^2=40=6^2+2^2+4^2 \implies x=6,y=-2, z=4$

$x^2+y^2=20=4^2+2^2+6^2\implies x=4,y=2,z=6$

---

$a-b=4. b-c=2. a-c=6$

$a=b+4$

$c=b-2$

---

$b=7, a=11, c=5$

$b=13, a=17. c=11$

$b=19, a=23, c=17$

$b=43, a=47, c=41$

---

$a^2+b^2+c^2-ab-ac-bc=28$

$b^2+(b+4)^2+(b-2)^2-b(b+4)-(b+4)(b-2)-b(b-2)=28$

$b^2+b^2+8b+16+b^2-4b+4-b^2-4b-b^2+2b-4b+8-b^2+2b=28$

$8b+16-4b+4-4b+2b-4b+8+2b=28$

$28=28$ regardless of $b$.

So select $b$ prime where $(b+4)$ and $(b-2)$ are prime. Open question how many such there are.