Can u gve an example of its existense?

Question

2^(n) = t^(m) - w^(m)
where, n =/= m.
like,2^(4) = 5^(2) - 3^(2)
give an example for
2^(4) = t^(3) - w^(3)
where "t" & "w" are natural.

Answer 1

2^4 = 16 = t^3 - w^3 is impossible in integers.This can be seen without a formal proof, with just the first few integers.
2^3 - 1^3 = 7
3^3 - 2^3 = 19
Already, the difference between t and w is only 1, and we have a number greater than 16 and they just keep getting larger.
e.g. 4^3 - 3^3 = 37 ... etc.

Here is a proof that 2^n = t^3 - w^3 is impossible for all integers n.
The general case for all m, as in your first equation, is out of my reach.

Let 2^n = t^3 - w^3
= (t - w)(t^2 + tw + w^2) ... (Equation 1)
This shows that 2^n is a product of 2 integers.

Now 2^n is always even. Therefore, t and w must be either both odd or both even.

1) Let t and w be both odd.
In (Equation 1), (t - w) will be even, but (t^2 + tw + w^2) will be odd.
But 2^n cannot be a product of an odd number, unless that number is 1.
But (t^2 + tw + w^2) can never equal 1.
This proves that t and w cannot both be odd.

2) Let t and w be both even.
So let t = 2p and w = 2q.
Substituting in 2^n = t^3 - w^3 gives :
2^n = (2p)^3 - (2q)^3 = 2^3(p^3 - q^3)
Dividing through by 2^3 gives :
2^(n-3) = p^3 - q^3

So first we had 2^n as a difference of 2 cubes.
Now we have a power of 2 that is less than 2^n, also being a difference of 2 cubes.
Continuing like this, we would have to have p and q as both even numbers and a similar equation would ensue, but being a lesser power of 2 than the previous one.
This could go on forever, but it can't, because there must be a minimum value of 2^1.
This proves that t and w cannot be both even, and thus concludes the proof that the original equation is impossible.

Answer 2

for m=1,2 of course there are solutions but for n > 2, not if natural.

We can't have one of t, w even since t-w would be odd and t-w divides t^(m)-w^(m). Both t,w even is ruled out for the same reason.
So both t,w odd and gcd(t,w)=1

Clearly m must be prime. For example, consider 2^n = t^6-w^6 = (t^3+w^3)(t^3-w^3). We would have to have t^3-w^3 = 2 which is impossible since obviously both factors must be powers of 2, and the 2nd factor is less than the first and one of them is only divisible by 2, since if we assumed 4 divides both factors then 4 divides the sum = 2t^2, but t is odd.

Now consider 2^n = t^3-w^3 = (t-w)((t-w)^2 +3tw)
If t-w is a power of 2 then the second factor is odd, so it is not possible for m=3

In general t^(m)-w^(m)/( t-w) will be an integer.

And the factors (t-w) and (t^(m)-w^(m))/( t-w) are co-prime if t and w are co-prime, which they must be, since if they weren't and the equation was true we would have a odd prime dividing 2^n since t,w are both odd.

Answer 3

No such t and w exists for the given condition since the difference between the smallest such combination say 3^3-1^3 itself is 26 and also 2^3-1^3 is obviously 7 which is not 16.Answer is no example for t and w can be given.