WakeUp wrote: A triple $(a,b,c)$ of positive integers is called quasi-Pythagorean if there exists a triangle with lengths of the sides $a,b,c$ and the angle opposite to the side $c$ equal to $120^{\circ}$. Prove that if $(a,b,c)$ is a quasi-Pythagorean triple then $c$ has a prime divisor bigger than $5$. $c^2 = a^2 + b^2 + ab;$ We can make, that $(a, b) = 1$. Let's watch : 1); mod 2: $a$ $b$ $c$ 1 1 1 It isn't divide on 2. 2); mod 3: $a$ $b$ $c$ 1 1 0 1 2 1 2 2 0 It divide on 3 - if $a \equiv_{3} b$. Let's watch: $a^2 + b^2 + ab \equiv_{9} 3ab + (a - b)^2 \equiv_{9} 3ab \not\equiv_{9} 0$! 3); mod 5: $a$ $b$ $c$ 1 1 3 1 2 2 1 3 3 1 4 1 2 2 2 2 3 4 2 4 3 3 3 2 3 4 2 4 4 3 It isn't divide on 5. That's all! !