## Task 1 - 150921

Klaus has to calculate $4^{2}-3^{2}$ for a homework assignment. He notices that the result, 7, is equal to the sum of the two numbers used, 4 and 3. When he checks his discovery with the numbers 10 and 11, he finds that $11^{2}-10^{2}=21=11+10$ as well.

Determine all pairs $(a, b)$ of natural numbers with $a>b$, for which the (positive) difference of the squares of the two numbers is equal to the sum of the two numbers!
According to the problem statement, all pairs ( $a, b$ ) with $a, b$ natural and $a>b$ are to be determined, for which $a^{2}-b^{2}=a+b$ holds.

Now, by a binomial formula, $a^{2}-b^{2}=(a+b)(a-b)$, and therefore the required property is equivalent to $(a+b)(a-b)=a+b$.

Since $a, b$ are natural and $a>b$, $a+b \neq 0$. Thus, the stated property is also equivalent to $a-b=1$, i.e., the given condition is satisfied precisely by the pairs $(a, b)$ of natural numbers for which $a$ is 1 greater than $b$.

Solution adopted from $[5]$