show that if both x and y are odd , then there is no integer z such that x²+ y²= z².
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Answers
Answer:
Answer:
Let the two odd positive no. be x = 2k + 1 and y = 2p + 1
Hence, 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
clearly, notice that the sum of square is even the no. is not divisible by 4
hence, if x and y are odd positive integer, then x2 + y2 is even but not divisible by four.
Step-by-step explanation:
We know that any odd positive integer is of the form 2q+1, where q is an integer.
So, let x=2m+1 and y=2n+1, for some integers m and n.
we have x
2
+y
2
x
2
+y
2
=(2m+1)
2
+(2n+1)
2
x
2
+y
2
=4m
2
+1+4m+4n
2
+1+4n=4m
2
+4n
2
+4m+4n+2
x
2
+y
2
=4(m
2
+n
2
)+4(m+n)+2=4{(m
2
+n
2
)+(m+n)}+2
x
2
+y2=4q+2, when q=(m
2
+n
2
)+(m+n)
x
2
+y
2
is even and leaves remainder 2 when divided by 4.
x
2
+y
2
is even but not divisible by 4.