## Task 1 - 060921

Give four different pairs $(a, b)$ of positive integers such that the difference of the squares of the two numbers in each pair is 105!

(Any two pairs $(a, b)$ and $(b, a)$ are considered not different from each other.)
From the problem statement, the following equation is derived:

$$
105=a^{2}-b^{2}=(a+b) \cdot(a-b) \quad \text { 3rd Binomial Formula }
$$

Now we substitute: $x:=a+b$ and $y:=a-b$.

Thus, it follows that: $a=x-b=y+b$. Therefore, we get:

$$
b=\frac{x-y}{2}
$$

and

$$
a=\frac{x+y}{2}
$$

105 has the divisors: $1,3,5,7,15,21,35,105$. Therefore, the following possible pairs $(x, y)$ exist with $x \geq y$, since $a+b \geq a-b$ if $a, b \in N:(105,1),(35,3),(21,5),(15,7)$.

Now we back-substitute using equations (1) and (2):

$$
\begin{array}{rr}
\text { Case 1: } & x=105, y=1 \Rightarrow a=53, b=52 \\
\text { Case 2: } & x=35, y=3 \Rightarrow a=19, b=16 \\
\text { Case 3: } & x=21, y=5 \Rightarrow a=13, b=8 \\
\text { Case 4: } & x=15, y=7 \Rightarrow a=11, b=4
\end{array}
$$