## Task $8 / 79$

Determine all prime numbers $a, b, c, d$ for which the following equations hold: $a+b=c$ and $2a+b=d$.
Since $d=2a+b>2a>2$, $d$ must be odd. Since $2a$ is an even number, it follows that $b$ is odd. Given $c=a+b>b$, $c$ is also odd, and thus $a=c-b$ is even. Therefore, $a=2$.

Of the prime numbers $b, b+2=c$, and $b+4=d$, exactly one is divisible by 3. This can only be the smallest of the three prime numbers. Therefore, $b=3$.

This results in $c=5, d=7$. According to the solution path, $(2, 3, 5, 7)$ is the only solution quadruple.