Math, asked by IƚȥCαɳԃყBʅυʂԋ, 7 months ago

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

Answered by Anonymous
4

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.

Answered by Mrsekhar
1

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.

Hope it helps you

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