2^x 3^y = z^2 1 where x, y, and z are nonnegative integers We can see that 1 is a congruence modulo 8 3^y = z^2 (mod 8) 2 When z=1 and and y=2 we have 3^2 = 1 (mod 8) 3 and by Fermat's theorem So this means that y must be even As y is even, we can write y = 2k for some integer k So we have 3^2k 2^x = z^2 z^2 3^2k = 2^xConsider x 3 y 3 z 3 − 3 x y z as a polynomial over variable x Find one factor of the form x^ {k}m, where x^ {k} divides the monomial with the highest power x^ {3} and m divides the constant factor y^ {3}z^ {3} One such factor is xyz Factor the polynomial by dividing it by this factorA classic way to prove inequalities is using AMGM inequality But my approach is different Here's my proof According to an algebraic identity, mathx^3 y^3 z
If X Y Z 0 Then Prove That X3 Y3 Z3 3xyz Youtube
