Inspired by Munem

Suppose that such a solution exists: $b^2 - nab + a^2 = 0$ .

Then, the quadratic equation gives us $b = \frac{ na \pm \sqrt{ n^2a^2 - 4 a^2 } } { 2 } = a \frac{ n \pm \sqrt{ n^2 - 4 } } { 2}$ .

Hence, we require $n^2 - 4$ to be a perfect square.

Suppose that $n^2 - 4 = m^ 2$ . Then, $n^2 - m^2 = 4$ , so $(n-m)(n+m) = 4$ .
Since $n-m , n+m$ are integers with the same parity, and are factors of 4, we must have $n -m = n+m = 2$ .
Thus $m = 0, n = 2$ .

However, we require $n \geq 3$ , so there are not solutions.

Note: When $n =2, m = 0$ , we have the positive integer solutions $a= b$ .

let us assume $a = kb$

$( n )( kb )( b ) =( kb )^2+( b^2 )$

$( b^2)( nk ) = [ ( b^2 )[ ( k^2 )+1 ]$

$nk =[ ( k^2 )+1 ]$

$n = [ ( k^2 )+1 ]$

$n = k + 1/k$

$n$ cannot have an integer value except for when $k =1$

but then $n=2$

but $n>=3$

• therefore the triple doesn't exist

let g=gcd(a,b).Then ,write a=gx and b=gy.

then it follows that,nxy= $x^2+y^2$ .Now,x and y both divides $x^2+y^2$ .Since,x and y are coprime ,we have x=y=1. That implies n=1+1=2.So,no solutions exist.