3b^2 = a^2 + c^2 , a,b, and c are rational

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someonekicked - 27 Nov 2006 08:13 GMT

HEllo all!

I want to find 3 rational numbers a,b and c such that 3b^2 = a^2 + c^2.

I am not sure where to begin!
I wrote a program in C++ to search for such numbers if a,b, and c were
positive integers, (I searched if a and c were between 1 and 10000).
But I did not find any.

Any suggestions?

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Giovanni Resta - 27 Nov 2006 08:34 GMT

> HEllo all!
>
[quoted text clipped - 6 lines]
>
> Any suggestions?

It is very easy to prove that there are no solutions.
Just consider the numbers mod 3, etc.etc.

g.

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Pubkeybreaker - 27 Nov 2006 14:05 GMT

> HEllo all!
>
[quoted text clipped - 6 lines]
>
> Any suggestions?

Hint: The left hand side is divisible by 3. What criteria must a & c
satisfy
so that a^2 + c^2 is divisible by 3?????

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someonekicked - 28 Nov 2006 10:31 GMT

I am looking for a,b, and c rational numbers! not integers..

if a = p1/p2, c = q1/q2, and b = s1/s2 where p1,p2,q1,q2,s1 and s2 are
integers,

Then after some calculation..

(p1q2s2)^2 + (p2q1s2)^2 = 3 s1^2.

So now we are back to the problem like if a,c and b were integers.

if a = c, then 3b^2 = 2a^2 certainly does not have solution since 3b^2
has "an odd number" of 3 factors in its prime factorization, while 2a^2
has an even numbers of 3 factors.

if a != c, I am still stuck here!

> HEllo all!
>
[quoted text clipped - 6 lines]
>
> Any suggestions?

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Pubkeybreaker - 28 Nov 2006 17:40 GMT

> I am looking for a,b, and c rational numbers! not integers..

Just clear denominators and you are led back to the
original equation... (as you point out)

> if a = p1/p2, c = q1/q2, and b = s1/s2 where p1,p2,q1,q2,s1 and s2 are
> integers,
[quoted text clipped - 4 lines]
>
> So now we are back to the problem like if a,c and b were integers.

You did not take the hint. The left side is 0 mod 3.
The squares mod 3 are 0,1, Thus if a or c are not both 0 mod 3 then
the rhs does not sum to 0 mod 3. Hence a,c must both be divisible by
3.
Now, 9 divides the rhs, but only 3 divides the left hand side unless
.........

Finish it.

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Math Forum / Mathematics / Undergraduate Math / November 2006

3b^2 = a^2 + c^2 , a,b, and c are rational