Problem 10.3. Anton wrote three natural numbers $a, b$, and $c$ on the board. Ira drew three rectangles $a \times b, a \times c$, and $b \times c$ on the board. It turned out that the difference in the areas of one pair of rectangles is 1, and the difference in the areas of another pair of rectangles is 49. What can $a+b+c$ be? List all possible options.
Answer: 16.

Solution. Without loss of generality, we assume that $1=ac-ab=a(c-b)$, then $a=1$, $c=b+1$. Thus, the numbers written on the board were $1, b, b+1$.

Notice that either $b(b+1)-b \cdot 1=49$, or $b(b+1)-(b+1) \cdot 1=49$.

In the first case, we get $b^{2}=49, b=7$, and $a+b+c=1+b+(b+1)=16$.

In the second case, we get $b^{2}=50$. Such natural $b$ does not exist.