Question

If m and n are odd positive integer the m2+n2 is even , not divisible by 4 justify

Answer 1

Supposing m and n to be odd integers, it can be said that m^2 and n^2 each would each be odd positive integers as well. To represent numerically, let's take m = 2p + 1, and n = 2q + 1. Then, m^2 + n^2 = (2p + 1)^2 + (2q+1)^2 = (8a + 1) + (8b + 1) (Since, the square of any odd positive integer can be represented as such, the proof can be answered in a different question if need be) = 8(a + b) + 2 Now, this sum would be indivisible by the 4 (even though the first term of the expression is a multiple of 8) as the second term would always give off a remainder of 2.

Answer 2

Step-by-step explanation:

Since m and n are odd positive integers, so let m = 2q + 1 and n = 2p + 1 ,

•°• m² + n² = ( 2q + 1 )² + ( 2p + 1 )² .

= 4( q² + p² ) + 4( q + p ) + 2 .

= 4{( q² + p² + q + p )} + 2 .

= 4y + 2 , where y = q² + p² + q + p is an integer .

•°• m² + n² is even and leaves remainder 2, when divided by 4 that is not divisible by 4.

Hence, it is solved

THANKS

#BeBrainly.