Prime Triangle

Draw a line parallel to BC, from D meeting AB at P. The quadrilateral PDCB is cyclic and so, BD^2=PD ×BC+DC^2,
PD=√3×AD, BC=√3×AC and DC=2, being the only even prime number as the difference between AC and AD, the two odd prime numbers. So, BD^2=3AD×AC + 4,
(BD+2)(BD - 2)=3AD×AC The right side are all prime numbers and BD being greater than AC, we have, BD + 2 = 3AD,
and BD - 2= AC, Subtracting 2nd from 1st, 3AD - AC=2AD - 2 =4, or AD=3. BD=3AD - 2= 9 - 2= 7.

2 is the only one even Prime. Only Even + Odd = 0dd .... $\color{#3D99F6}{(***)}$ .
AB=AC=AD + DC. AC is prime and so odd, AD is prime and so odd, AD > DC, also implies AC > DC.
We know $\color{#3D99F6}{(***)} .\ \ So\ \ \color{#3D99F6}{AD=2}. \ a\ must.$
AD + 2 =AC=AB. So AB, AC are twin prime.
$BD=\sqrt{AB^2 + AD^2 - 2*AB*AD*Cos120}......Cos\ Rule....\color{#3D99F6}{(*)}$
So we investigate twin primes.
Start with (3,5).
Applying Cos Rule as in $\color{#3D99F6}{(*)}\ \ BD=\sqrt{5^2 + 3^2 - 2*5*3*( - \frac 1 2) } \ =\ 7$ a prime not used.
BD= 7 .

Relevant wiki: Solving Triangles - Problem Solving - Medium

Since $BA = AC = AD + DC$ , and all these lengths are prime, $DC$ is even or equal to $2$ , the only even prime available. Then let $AD = x$ , so $BA = AC = x+2$ . We can apply the cosine rule for $\triangle ABD$ :

$BD^2 = (x + 2)^2 + x^2 - 2(cos 120^\circ)(x(x+2)) = 3x^2 + 6x + 4$

Since $BD$ is also prime in length, so we let $BD = x +2n$ for some integer $n$ .

$(x + 2n)^2 = 3x^2 + 6x + 4$

$0 = 2x^2 + (6-4n)x + 4(1 - n^2)$

$0 = x^2 + (3-2n)x + 2(1 - n^2)$

Solving for $x$ , we will get: $x = \dfrac{(2n-3) \pm \sqrt{(3-2n)^2 - 8(1-n^2)}}{2} = \dfrac{(2n-3) \pm \sqrt{12n^2 - 12n +1}}{2}$

The square root of the quadratic discriminant must be an integer, so the discriminant equals to $m^2$ for some integer $m$ .

$12n^2 - 12n + 1 = m^2$

$12(n^2 - n) = m^2 - 1 = (m - 1)(m + 1)$

By factorization to its near square root, the left hand side will have two factors, whose absolute difference is $2$ .

Then we can rewrite the equation as:

$(3n)(4(n-1)) = (4n)(3(n-1)) = (2n)(6(n-1)) = (6n)(2(n-1)) = (n)(12(n-1)) = (12n)(n(n-1)) = (12)(n^2 - n) = (m-1)(m+1)$

Starting from the first terms, the factors $3n$ and $4(n-1)$ will equal to $m-1$ or $m+1$ . Hence, the absolute difference between factors will equal to $2$ :

$|3n - 4(n-1)| = 2$

Thus, $4 - n = 2$ or $4 - n = -2$ . Then $n$ = $6$ or $2$ .

By using the same method, for terms $(4n)(3(n-1))$ , we will get $n$ = $-1$ or $-5$ .

For $(2n)(6(n-1))$ , $n$ = $1$ or $2$ .

For $(6n)(2(n-1))$ , $n$ = $0$ or $-1$ .

For other terms, however, $n$ can never be an integer.

Therefore, the possible value of $n$ is an element in the set {-5, -1, 1, 2, 6}. (Note that $n$ can't be zero because of different prime numbers.)

By substituting values of $n$ , the value of $x$ can be evaluated as followed:

$n = -5; x = 3\ or \ -16$ $n = -1; x = 0\ or \ -5$ $n = 1; x = 0\ or \ -1$ $n = 2; x = 3\ or \ -2$ $n = 6; x = 14\ or \ -5$

Out of these outcomes, $3$ is the only positive prime solution for $x$ , yet the only applicable value of $n$ is $2$ . That is, $BD = 3 + 2\times 2 = \boxed{7}$ .