Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

Solve

SOLVE
a,b integers

Last edited by tony123 (2007-12-03 22:19:42)

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Solve

Edit: Seems like this solution is flawed

As the disciminant is not a perfect square, there exist no integer solutions.

Last edited by Identity (2007-12-13 01:01:10)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Solve

Oopsy, you made a mistake in the second-last line: it should be

Alternatively, we can solve the equations by inspection:

Last edited by JaneFairfax (2007-12-04 00:23:16)

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Solve

Oh math! Thanks for spotting that Jane. I'll try another solution:

We start by proving that a and b must both be non-negative perfect squares:

As a must be an integer, so must

and therefore

, so b is a perfect square.

We can similarly prove the same thing for a.

Now, both must be non-negative because they are square rooted and real + imaginary =/= real, as long as the imaginary part =/= 0. By observing 11 and 7 are not perfect squares, we can prove a,b=/=0.

So we are restricted to the values {(a,b): (1,1),(1,4),(1,9),(4,1),(4,4),(4,9),(9,1),(9,4),(9,9)}, and through checking cases we find that the only solution is {(a,b): (9,4)}

tony123

Member

Registered: 2007-08-03

Posts: 229

Re: Solve

this is  other way

 

----1

from 1

Last edited by tony123 (2007-12-04 06:02:36)