**Question:**

Prove that there cannot be two integers m and n so that the following equation is satisfied: $ 2m ^ {2} + n ^ {2} = 2007 $.

**Answer**

Assuming there are two integers $ m $, $ n $ so: $ 2m ^ 2 + n ^ 2 = 2007 $ (*)

Since $ 2m ^ 2 $ is even, we deduce $ n ^ 2 $ must be an odd number or $ n $ must be an odd number, $ n = 2a + 1 $ ($ a \in \mathbb {Z} $).

(*) $ \Leftrightarrow 2m ^ 2 + ( 2a + 1) ^ 2 = 2007 \Rightarrow 2m ^ 2 + 4a ^ 2 + 4a + 1 = 2007 $

$ \Rightarrow 2m ^ 2 + 4a ^ 2 + 4a = 2006 $

Because $ 4a ^ 2 + 4a $ divisible by $ 4 $ and $ 2006 $ not divisible by 4.

Inferred $ 2m ^ 2 $ undivided by $ 4 $ or $ m ^ 2 $ undivided by $ 2 \Rightarrow $ $ m ^ 2 $ is an odd number. Put $ m = 2b + 1 $, we have: $ 2 (2b + 1) ^ 2 + 4a (a + 1) = 2006 $

$ \Rightarrow 8b ^ 2 + 8b +2 + 4 (a + 1 ) a = 2006 $

$ \Rightarrow 8b (b + 1) + 4a (a + 1) = 2004 $.

It’s easy to see that $ 8b (b + 1) $ is divisible by $ 8; 4a (a + 1) $ is also divisible by $ 8 $ and 2004 is not divisible by $ 8 $. This conflict gives us fuck.