What is the right value for a^2 + ab + b^2 can be a perfect square?

Question

we all know that any value given to a^2 + 2ab + b^2 is always a perfect square, theres an answer to my question but I forgot what
please give me! tnx

oh I need a constant please! tnx!

also a or b is not equal to zero

Answer 1

Interesting question. I didn't get anywhere with it. Some failed attempts are below.

EDIT: Good point above. b = -a is always a solution, unless that contradicts some assumption in the problem, like that they're both positive.

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If there's an answer at all, there's one with a and b relatively prime. So let's assume they are.

Rewrite it as (a+b)^2 = c^2 + ab

If a and b are both odd, then so is c. And ab is congruent to 3 (mod 4), since c^2 is congruent to 1.

If one of them is even -- say a -- then it has to in fact be divisible by 4. And c is odd. (Again, this is a result of looking at both sides mod 4.)

OK. This isn't getting anywhere ...

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How about brute forcing it?

a^2 + ab + b^2 = (a + b/2)^2 + 3b^2/4

Nope. Trying to factor that into two terms, each of which is a perfect square at least of rational numbers, fails miserably. Multiplying both sides by 4 gives a RHS of (2a+b)^2 + 3b^2, which doesn't particularly cause more joy.

Answer 2

A fraction answer comes to mind
where a = 2
and b = 1/2
In that case your equation comes to a value of 5 and a quarter which is the square of 1 and a half!!

(3 hours later)

I just got an answer by trial and error a = 3 and b = 5 or we can have any whole number values in the ratio of 3 : 5 such as 6 & 10 or 9 & 15 and so on.

Answer 3

If a^2=-ab or a=-b or b^2= -ab or b= -a ,then a^+ab+b^2 can be a perfect square.In that case either a^2+ab or ab+b^2 will cancel each other and only a^2 or b^2 will remain
But I myself hate this answer.Just for answer's sake I am giving this answer

Answer 4

(a+b)^2

Answer 5

This is not an intelligible question.