Prove that if x and y are odd positive integers then x square + y square is even but not divisible by 4
Answers
Answered by
2
Let the two odd numbers be
(2a+1) & (2b+1) because if we add 1 to any even no. it will be odd.
x²+y²
=>(2a+1)²+(2b+1)²
=>(4a²+4a+1)+(4b²+4b+1)
=>4(a²+b²+a+b)+2
4 Is not a multiple of 2 it means clearly that 4 is not multiple of x²+y² , so x²+y² is even but not divisible by 4.
Hence proved.
Hope this helps you
Please mark it as brainliest
(2a+1) & (2b+1) because if we add 1 to any even no. it will be odd.
x²+y²
=>(2a+1)²+(2b+1)²
=>(4a²+4a+1)+(4b²+4b+1)
=>4(a²+b²+a+b)+2
4 Is not a multiple of 2 it means clearly that 4 is not multiple of x²+y² , so x²+y² is even but not divisible by 4.
Hence proved.
Hope this helps you
Please mark it as brainliest
Answered by
0
Step-by-step explanation:
we know that any odd positive integer is of the form 2q+1 for some integer q.
so, let x = 2m+1 and y = 2n+1 for some integers m and n.
•°• x²+y²= ( 2m+1)²+(2n+1)²
==> x²+y²=4(m²+n²)+4(m+n)+2
==> x²+y²= 4{(m²+n²)+(m+n)} +2
==> x²+y² = 4q +2 , where q= (m²+n²)+(m+n)
==> x²+y² is even and leaves remainder 2 when divided by 4.
==> x²+y² is even but not divisible by 4.
HENCE PROVED✔✔
Similar questions