2016_Question_3

2016_Question_3

Solution:
First we note that we can rearrange the equation and complete the square in n to get:
(n-3)^2-m^2=m-1

Now we can notice that if we have two squares, (n-3)^2 and m^2, (assuming that they are not equal), the difference has to be \geq 2m+1 because (m+1)^2-m^2 = 2m+1.

But 2m+1 > m-1 for our values of m so thus the difference between two squares must be 0, which means that m-1=0 \longrightarrow m=1.

Solving with m=1 we get that:
(n-3)^2=1, which is true when n=4 or n=2.

Thus our only possible pairs are (1, 4) and (1, 2).

:blue_heart: :large_blue_diamond: :small_blue_diamond: Team Blue!!! :small_blue_diamond: :large_blue_diamond: :blue_heart:

1 Like

:blue_heart: :large_blue_diamond: :small_blue_diamond: FANTASTIC :grin:!!! :small_blue_diamond: :large_blue_diamond: :blue_heart: