Possible-number-of-order-pairs-satisfy-16-x-2-y-16-y-2-x-1-is-

Question Number 171226 by SLVR last updated on 10/Jun/22

P o s s i b l e n u m b e r o f o r d e r p a i r s

s a t i s f y 16^{x^{2} + y} + 16^{y^{2} + x} = 1 i s

Commented by SLVR last updated on 11/Jun/22

c a n a n y o n e m e p l e a s e

Commented by SLVR last updated on 10/Jun/22

i m e a n p a i r s (x, y) i s h o w m a n y ?

k i n d l y h e l p m e o u t

Commented by mr W last updated on 11/Jun/22

2^{4 (x^{2} + y)} + 2^{4 (y^{2} + x)} = 1

4 (x^{2} + y) = 4 (y^{2} + x) = - 1

\Rightarrow x = y = - \frac{1}{2}

Commented by SLVR last updated on 11/Jun/22

S o k i n d o f y o u p r o f e s s o r .

G r e a t l y t h a n k y o u

Facebook Tweet Pin