Prove this trick?

Question

prove whether it is right or wrong
It is impossible to write
a^(n)=b^(n+1) except for a=b=1 & 0 for natural solutions of "a" & "b".

Answer 1

False if a= b^2 else

this is true(with one additionall exeption a= b^2 and n= 1 satisfiles the condition so a should not be b^2)

if (a)^n = b^(n+1)

then b = (a/b)^n
deviding both sides by b^n
now let a and b hace gcd t

a= xt
b = yt
and x and y are co-primes

yt = (x/y)^n

LHS is integer and RHS is not an integer as x y are coprimes.

so a contradiction unless x = y
in tnis case yt = 1
so t and y and x all are 1
so a=b=1 holds

a=b=0 is a trivial case as both are zero
so the statement is correct when a is not multiple of b

in case a is multiple of b say

a= bt
b = t
so when a = b^2
now a^n = b^2n
so a = b^2

for all other cases it is true

Answer 2

Your question doesn't make it a necessary condition that a = b for all solutions (a, b), but just says that the only possible solutions are a = b = 1 & 0. I think it was meant to be worded differently, but in this original form, the question is false since there are an infinite number of solutions if it is permitted that a and b are not equal to one another.

[To provide a counterexample, n = 1 gives a = b^2, for which there are an infinite number of solutions (a, b).]

To not evade the question entirely, I'll assume that the statement was written in such a way that it only deals with solutions where a = b: "The only natural solutions (a, b) of a^(n)=b^(n+1) such that a = b are (0, 0) and (1, 1)."

a^n = b^(n+1)

a^n = a^(n+1)

a^(n+ 1) - a^n = 0

a^n (a - 1) = 0

a = 0 or a = 1

Therefore, the statement (in a different form) is true.

Answer 3

another way to write the equation would be (a/b)^n=b. one solution is a=8 b=4 n=2. another would be a=27 b=9 n=2.
now lets consider n=3. and b= 8 which is a cube of 2. so a= 16 b=8 n=3. all we have to do is keep on changing a to get (a/b)^n as b. so we have an infinite number of solutions. so the assumption in the question is false.

Answer 4

wrong..
lets say a= 5
b = 6
and n=5 then

5^(5)==6^(6);

that is 5 exor 5 is 101^ 101 = 000
6^6 again 0..

so the following statement is wrong

It is impossible to write
a^(n)=b^(n+1) except for a=b=1 & 0 for natural solutions of "a" & "b".

Answer 5

from your assumption

a^n = b^(n+1)

= a = (b^ (n+1)) ^ (1/n)

= a = (b ^ ( n*1/n + 1*1/n ) )

= a = (b ^(1 + 1/n))

= a = (b*b^(1/n))

= a / b = b^(1/n)

now a = 8 and b = 4
= 2 =2^(1/n)
n= 2

for your
ans
a=8 b =4 n =2

8^2 = 64

4^(2+1) = 4^3 =64

Answer 6

Not true...

Let a = 4, b = 2, n = 1

(By the way, I found this because I assumed it was right but ran into a problem proving it. It turns out the problem is that there are counterexamples, heh heh.

Answer 7

a^n = b^(n+1)
a^n = b*b^n
if a=b=1 or a=b=0
then a^n=b^(n+1) as 0^(anything) & 1^(anything) both are 1.

otherwise
you cannot prove whether it is right or not.
eg a=4; b=2;
it is only right for n=1;

additionally you could say,
a^n=b^(n+1)
if sqrt(a)=b & n=1;

Answer 8

AREA 51 might come looking for you

Answer 9

(a/b) ^n = b

take a = b*b and n = 1 and you will have a lot of solutions
and hter are many many more

Answer 10

question in itself is not self explanatory
wat vaue does a and b can have ?
is there any relation b/w a & b?
useless question