LIII OM - III - Task 1

Determine all such triples of natural numbers $ a $, $ b $, $ c $, such that the numbers $ a^2 +1 $ and $ b^2 +1 $ are prime and
Without loss of generality, we can assume that $ a \leq b $. Let $ p = b^2 +1 $. Then the number $ (c^2 +1) - (b^2 + 1) = (c - b)(c + b) $ is divisible by $ p $. From the given equation in the problem, it follows that $ b < c $. Furthermore,

Thus $ c - b < p $ and $ c + b < 2p $. Since $ p | (c - b)(c + b) $, it must be that $ c + b = p $, which means

The numbers $ a = b = 1 $ do not satisfy the conditions of the problem; we can therefore assume that $ b > 1 $. In this case, $ p $ is an odd prime number. From the relationship (1), it follows that $ c $ is odd, which in turn proves the evenness of the numbers $ c^2 + 1 $ and $ a^2 + 1 $. The number $ a^2 + 1 $ is prime, so $ a^2 + 1 = 2 $, from which $ a = 1 $. Thus, the given equation in the problem takes the form

Combining equations (1) and (2), we get $ 2b^2 + 2 = b^2(b - 1)^2 + 2b(b - 1) + 2 $, which simplifies to $ 2b = b(b - 1)^2 + 2b - 2 $. From this, $ b $ is a divisor of the number $ 2 $, and since $ b > 1 $, it must be that $ b = 2 $. This implies $ c = 3 $.
It remains to state that the triplet $ (a,b,c) = (1,2,3) $ satisfies the conditions of the problem.
Releasing the assumption $ a \leq b $, we obtain two triplets $ (a,b,c) $ that satisfy the conditions of the problem. They are: $ (1, 2, 3) $ and $ (2, 1, 3) $.