Question

We have:
x^2+y^2=625 and GCD(x,y)=1
how can you prove that (x,y) is (even,odd) or (odd,even)

Answer 1

let me begin by proving that x and y both are not even numbers :
Because GCD(x,y)=1
:)

Answer 2

So now we need to prove that they are not odd in the same time

Answer 3

hint : square of an odd number is odd,
and sum of two odd numbers is even

Answer 4

good, what about the square of even number ?

Answer 5

if it is an even number, then definitely it has a prime factor of 2.
so its square also will have a prime factor of 2 right ?

Answer 6

a is even => a^2 is even
a is odd => a^2 is odd

Answer 7

that's right

Answer 8

even + even = even
odd + odd = even

so, left hand side evaluates to even number when both x and y are even or both are odd
QED

Answer 9

625 is not even so a^2+b^2 is not even too => (a,b) != (even,even)

Answer 10

also (a, b) != (odd, odd) for the same reason

Answer 11

But 625 is odd, isn't it ?

Answer 12

Ah Yes, odd+odd != even that's right

Answer 13

@ganeshie8 good ideas, thinking of the sum of even and numbers is great way to prove that, actually i was looking for GCD part

Answer 14

even and odd **

Answer 15

looks ikram has some thing to use GCD to arrive at the same.... lets wait :)

Answer 16

hmm i was gonna say somthing but not sure of it nw

Answer 17

:)

Answer 18

ok it work even if gcd is not 1

Answer 19

I will give up things that give me the part of answer

Answer 20

yah... itt works in general. bugcd(x,y)=1 is required for right side to be a perfect square

Answer 21

ya.. i missed that :o

Answer 22

*odd perfect square

Answer 23

but no lol

Answer 24

20^2 + 15^2= 25^2

Answer 25

ugh (facepalm)

Answer 26

@ganeshie8 looking for an alternative to (facepalm) in Latex ?

Answer 27

lol latex is static i think... i gtg for lunch... brb

Answer 28

HH Compose something useful

Answer 29

Thank you people