LV OM - I - Task 7

Find all solutions to the equation $ a^2+b^2=c^2 $ in positive integers such that the numbers $ a $ and $ c $ are prime, and the number $ b $ is the product of at most four prime numbers.
There are three solutions $(a,b,c): (3,4,5), (5,12,13), (11,60,61)$. Let $a$, $b$, $c$ be numbers satisfying the conditions of the problem. Then

from the assumption that the number $a$ is prime, we obtain $c = b + 1$. Therefore, $a^2 = 2b + 1$, which implies that the number $a$ is odd.
Let $a = 2n + 1$. Then

if the number $n$ gives a remainder of 1 when divided by 3, then the number $a$ is divisible by 3. Therefore, it is composite, except for the case $n = 1$, which gives the solution $(a,b,c) = (3,4,5)$. If, on the other hand, the number $n$ gives a remainder of 2 when divided by 5, then the number $a$ is divisible by 5. Therefore, it is composite, except for the case $n = 2$. This case gives the second solution $(a,b,c) = (5,12,13)$.
If, however, the number $n$ gives a remainder of 1 or 3 when divided by 5, then the number $c$ is divisible by 5. Therefore, it is composite except for the case $n = 1$, which we have already considered.
There remains the case where the number $n$ gives a remainder of 0 or 2 when divided by 3, and a remainder of 0 or 4 when divided by 5. In this case, the number $b$ is divisible by $2 \cdot 6 \cdot 5 = 60$. On the other hand, the number $b$ is the product of at most four prime numbers, which implies that $b = 60$. This case leads to the third solution: $(a,b,c) = (11,60,61)$.