Why $a^{2}+b^{2}\neq c^{2}$ when $a=b=c=1$ doesn't violate Pythagoras' Theorem? My exercise is this:
An equilateral triangle whose side lengths are equal to 1. Observe that in this particular case, $a^{2}+b^{2}\neq c^{2}$. Explain why this doesn't violate the Pythagoras' Theorem.
I'm stuck with this exercise. Give some advice what I need to check before trying to answer this question.
Is there any exception for equilateral triangles due to side lengths equality?
 A: The Pythagorean theorem applies only to triangles with a right angle. 
A: When something violates an estabilished theorem, the reason is always that it doesn't respect its hypothesis. So, what does Pythagoras' Theorem say?
From Wikipedia
In mathematics, the Pythagorean theorem, also known as Pythagoras's theorem, is a relation in Euclidean geometry among the three sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. 
What is the hypothesis? Well, just one fact: "we are in a right triangle". So you have to ask yourself: am I in a right triangle?
A: All answers are ok, but on a sphere of radius $\frac{2}{\pi}$, you can have an equilateral triangle of sides $1$ with 3 right angles.
But the Pythagorean theorem only applies in euclidean geometry, not in spherical geometry.
So you may violate one or another hypothesis (this answer is just to illustrate that "a right angle" is not the only hypothesis of this theorem)
A: $a^2 + b^2 = c^2$ is only true in the case of a right triangle (with one angle of the triangle being 90 degrees). Since you are working with an equilateral triangle, all angles of the triangle are 60 degrees and thus, you do not have a right triangle.  
In such a case where you know you have an equilateral triangle and you want to know any angles or sides, the law of cosines (a more general form of the Pythagorean theorem) or the law of sines would work.
