AN ATTEMPT TO SOLVE TRIGO PROBLEM.

Details including whole numbers a, b, c

S =2*p p =2*q

Basic premise

Long unit C c =1

C > b > a (a + b) > c

We have to prove that S (in a^s + b^s = c^s) isn’t bigger than 2.

Proof: a =/= b* .1

A^2 + b^2 = 2*(b)^2 =c^2

2^.2*b =c

Continuation .2

According in “Co sinus’s Theory”

A^2 + b^2- 2*a*b*Cos© =c^2

When© has a right angle we get

A^2 + b^2 = c^2

` and the opposite

When a^2 + b^2 =c^2

Then C is a right angle.

According to Sinus Theory

A/Sin(A) = b/Sin(B) =c/Sin(C)

When C is a right angle and c=1

We get

A = Sin(A), b =Sin(B), c = Sin(C).

X^s=(X^p)^2

Y^s =(Y^p)^2

According to the trigonometry theory

When a triangle has a right angle

Sin^2(A)+Sin^2(B) = 1

A^2+b^2 =1

(a^p)^2 + (b^p)^2 =1

(a^(2*q))^2 + (b^(2*q))^2 = 1

We get:

s = 2, p =1, q = 0.5, or:

An equal on one side a^2, b^2, c^2.

And the ather side (a^p)^2, (b^p)^2, (c^p)^2.

Of the same function

Is a contradiction between the two sides.

Have been proved.

Matus Caspi

Yagur 30065

Israel