Question -
Answer -
Solution:
Let the two odd positive numbers x and y be 2k+ 1 and 2p + 1, respectively
i.e., x2┬а+ y2┬а=(2k + 1)2┬а+(2p + 1)2
= 4k2┬а+ 4k + 1 + 4p2┬а+4p + 1
= 4k2┬а+ 4p2┬а+4k + 4p + 2
= 4 (k2┬а+ p2┬а+k + p) + 2
Thus, the sum of square is even the number isnot divisible by 4
Therefore, if x and y are odd positiveinteger, then x2┬а+ y2┬аis even but not divisibleby four.
Hence Proved