Find all positive integers $L$, $M$, $N$ such that $L^2 + M^2 = \sqrt{ N^2 +21}$

Let $(L^2 + M^2) = A$. Then you have $A^2 = N^2 + 21$ i.e. $(A+N)(A-N) = 21 = 21 \times 1= 7 \times 3$. Now work out the different possibilities with the constraint that you are dealing with positive integers. Move your mouse over the gray area to find the solution.

The first one i.e. $21 \times 1$, gives us $A = 11$ and $N = 10$, while the second one i.e. $7 \times 3$ gives us $A = 5$ and $N = 2$. $A = 11$ cannot be expressed as sum of two squares since it is $3 \bmod 4$. $A=5$ implies that $L^2 + M^2 = 5$ which gives us $(L,M) = (2,1)$ or $(L,M) = (1,2)$. Hence, $(2,1,2)$ and $(1,2,2)$ are the only solutions.