If m and n are odd positive integers, then m² + n² is even, but not divisible by 4. Justify.✓
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Step-by-step explanation:
Given:-
m and n are odd positive integers
To find:-
Show that m^2+ n^2 is even, but not divisible by 4
Solution:-
Given that
m and n are odd positive integers
We know that
the general form of an odd positive integer = 2a+1
Let m = 2a+1
and n = 2b+1
m^2 = (2a+1)^2
=> m^2 = (2a)^2+2(2a)(1)+(1)^2
(Since (a+b)^2=a^2+2ab+b^2)
=> m^2 = 4a^2+4a+1---------(1)
and
n^2 = (2b+1)^2
=> n^2 = (2b)^2+2(2b)(1)+(1)^2
(Since (a+b)^2=a^2+2ab+b^2)
=> n^2 = 4b^2+4b+1---------(2)
On adding (1)&(2) equations then
=> m^2+n^2
=>(4a^2+4a+1)+(4b^2+4b+1)
=> 4a^2+4a+1+4b^2+4b+1
=> 4a^2+4a+4b^2+4b+2---------(3)
=> 2(2a^2+2a+2b+1)
=> 2c, Where c = 2a^2+2a+2b+1
=> It is the multiple of 2
=> It is in the form of the even number
=> m^2+n^2 is an even number------(4)
from (3) we notice that
each term is divisible by 4 except 2
so ,
The total expression does not divisible by 4---(5)
From (4)&(5)
We conclude that
"If m and n are odd positive integers, then
m^2 + n^is even, but not divisible by 4.
Hence, Proved.
Check:-
Let consider two odd positive integers 3 and 5
3^2 = 9
5^2 = 25
3^2+5^2
= 9+25
= 34
=> it is an even number and does not divisible by 4.
Verified the given relation.