Prove that △ABC is not a right triangle where BC=a^10, CA=b^100, AB=c^1000 and a, b, c are natural numbers. (my original quiz)?

Fermat's Last Theorem
No three positive integers i , j , and k satisfy the equation :
i^n + j^n = k^n , for any integer n > 2

Suppose ∠A = 90°
Then BC² = AC² + AB²
( a^10 )² = ( b^100 )² + ( c^1000 )²
a^20 = b^200 + c^2000
( a² )^10 = ( b^20 )^10 + ( c^200 )^10
Take k = a² , i = b^20 , j = c^200
Since a, b, c are natural numbers, i , j , k are also natural numbers.
Then k^10 = i^10 + j^10 , a contradiction with The Fermat's Last Theorem.

Similarly, both ∠B = 90° and ∠C = 90° are contradiction with The Fermat's Last Theorem.
Hence, △ABC is not a right triangle.
Q.E.D.