Let's determine the positive prime numbers $p>q>r$ for which

$$
p^{2}-(q+r)^{2}=136
$$
Solution. Let's transform the left side of the equality into a product, and decompose the number on the right side into prime factors:

$$
(p+q+r)(p-q-r)=17 \cdot 2 \cdot 2 \cdot 2
$$

The factors on the left side are built from the numbers on the right side. These can be: $17 \cdot 8, 34 \cdot 4$, or $68 \cdot 2$, or $136 \cdot 1$. From the condition $p+q+r > p-q-r > 0$ and $(p+q+r) + (p-q-r) = 2p$ is even, only the second and third cases can apply.

For

$$
\left.\begin{array}{l}
p+q+r=34 \quad \text { and } \\
p-q-r=4
\end{array}\right\} \Longrightarrow 2 p=38 \text { and } p=19
$$

furthermore, $q+r=15$, which is possible if and only if $q=13$ and $r=2$. This is thus a solution to the equation.

Finally, if

![](https://cdn.mathpix.com/cropped/2024_05_02_6bc6d5ed9c3d6596ff66g-1.jpg?height=111&width=731&top_left_y=622&top_left_x=689)

but this is not a prime number.

The solution to the problem is the triplet $19, 13, 2$.