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Prove that if x and y are both odd positive integers then x2 + y2 is even but not divisible by 4.
Prove that if x and y are both odd positive integers then x2 + y2 is even but not divisible by 4.
Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.
Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.
Let the two odd positive numbers x and y be 2k + 1 and 2p + 1, respectively
i.e., x2 + y2 = (2k + 1)2 +(2p + 1)2
= 4k2 + 4k + 1 + 4p2 + 4p + 1
= 4k2 + 4p2 + 4k + 4p + 2
= 4 (k2 + p2 + k + p) + 2
Thus, the sum of square is even the number is not divisible by 4
Therefore, if x and y are odd positive integer, then x2 + y2 is even but not divisible by four.
Hence Proved
Since x and y are odd positive integers, Then it should be in the form of 2x+1 (where x is a positive integer)Let x = 2m + 1 and y = 2n + 1 ( where m and n positive integers)Now x2 + y2= (2m + 1)2 + (2n + 1)2= 4m2 + 4m + 1 + 4n2 + 4n + 1= 4(m2 + n2 + m + n) + 2 ……. (1)from (1) we getx2 + y2 = 2 {m2 + n2 + m + n) + 1} = 2t (t = 2(m2 + n2 + m + n) + 1 is a positive integerTherefore it is clear that {tex}x^2+y^2{/tex} is an even number but not divisible by 4.
Let us consider two odd positive numbers be x and y where
x = 2p + 1 and y = 2q + 1
From question,
x2 + y2 = (2p + 1)2 +(2q + 1)2
= 4p2 + 4p + 1 + 4q2 + 4q + 1
= 4p2 + 4q2 + 4p + 4q + 2
= 4 (p2 + q2 + p + q) + 2
Form above result, we can conclude that x and y are odd positive integer, then x2 + y2 is even but not divisible by four.