Prove that 15x^2 -7y^2=9 has no integer solutions.
I have understood a few steps
15x^2 -7y^2=9 => y=3y1 => 5x^2-21y^2=3 => x-3x1 => 15x1^2-7y1^2=1 =>y1^2=2 mod 3
But this contradiction how???
If there is any other way plz tell.
Thanx
Prove that 15x^2 -7y^2=9 has no integer solutions.
I have understood a few steps
15x^2 -7y^2=9 => y=3y1 => 5x^2-21y^2=3 => x-3x1 => 15x1^2-7y1^2=1 =>y1^2=2 mod 3
But this contradiction how???
If there is any other way plz tell.
Thanx
I'm sorry that's a little hard to make out, presumably what it says is that if were solutions then and so but since is prime and we may conclude that and so and so we must have that but in the same spirit we see that this implies and so and since is prime and we may conclude that and so and so our equation becomes, after dividing both sides by , the following but this says that but it's evident that isn't a quadratic residue /
Hello, Learner248!
I am puzzled, too.
Prove that: has no integer solutions.
I have understood a few steps
But this contradiction how?
If there is any other way, plz tell.
We have: .
Since and are multiples of 3,
. . then must be a multiple of 3.
Hence, is a multiple of 3: .
The equation becomes: .
. . Divide by 3: .
Since and are multiples of 3,
. . then must be a multiple of 3.
Hence, is a multiple of 3: .
The equation becomes: .
. . Divide by 3: .
And I have no idea where they got:
. . It doesn't seem to be true.
I think I can finish the proof . . .
We have: .
Then: . .[1]
Since is an integer, then must be a multiple of 7.
. . That is: .
Substitute into [1]: .
. . which simplfies to: .
For integer values of will end in 2 or 7.
But squares must end in
There is our contradiction!