MHB Solve for Positive Integer Solutions

[anemone]

Find all values of $(a,\,b)$ where they are positive integers for which $\dfrac{a^2+b^2}{a-b}$ is an integer and divides 1995.

 

[Opalg]

[sp]If $\frac{a^2+b^2}{a-b} = k$, then $a^2+b^2 = k(a-b)$. Multiply by $4$ and complete the square, to get $(2a-k)^2 + (2b+k)^2 = 2k^2.$ We want to find factors $k$ of $1995 = 3\cdot 5\cdot 7 \cdot 19$ such that $2k^2$ is a sum of two squares. Now the only way that a number can be expressed as the sum of two distinct squares is if it has factors congruent to $1$ mod $4$. The only such factor in $1995$ is $5$. If we put $k=5$ then $2k^2 = 50 = 1^2 + 7^2$. Putting $2a-5=1$ and $2b+5 = 7$, we get the solution $(a,b) = (3,1).$ The only other solutions will occur through multiplying this basic solution by another factor of $1995.$ Those factors are $3,7,19,21,57,133$ and $399$. Thus there are eight solutions altogether namely $$(a,b) = (3,1),\ (9,3),\ (21,7),\ (57,19),\ (63,21),\ (171,57),\ (399,133),\ (1197,399).$$[/sp]

 

[anemone]

Bravo, Opalg! (Cool)(Clapping)(Sun) And thanks for participating!:)