there-is-an-ineterger-a-b-c-can-there-be-an-interger-as-a-n-b-n-1-c-n-2-n-is-a-interger-

Question Number 5152 by 1771727373 last updated on 23/Apr/16

t h e r e i s a n i n e t e r g e r a, b, c

c a n t h e r e b e a n i n t e r g e r a s

a^{n} + b^{n + 1} = c^{n + 2} (n i s a i n t e r g e r)

Commented by Yozzii last updated on 25/Apr/16

L e t n = 0 \Rightarrow 1 + b = c^{2}

L e t (a, b, c) = (0, 8, 3)

\Rightarrow 1 + 8 = 3^{2} \Rightarrow 9 = 9 .

Commented by Rasheed Soomro last updated on 06/May/16

If (a, b, c) = (0, 8, 3) then for n = 0

a^{n} + b^{n + 1} = c^{n + 2} \Rightarrow 0^{0} + 8^{1} = 3^{2}

Does 0^{0} = 1 ?

I think 0^{0} is indeterminate and

a is an integer other than 0 .

Commented by prakash jain last updated on 08/May/16

0^{0} is indeteminate only when it occurs as

a limit in that case that value of limits

depends on the function, in this case

\lim_{a \to 0} a^{0} = 1

Commented by Rasheed Soomro last updated on 08/May/16

This is knowledge for me . Thanks!