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How many solutions of n are possible if [ a , b , n ε N ]
a n = ( b + 1 ) n + ( a − 1 ) n ?
note 0 ∈ / N
The answer is 1.
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1 solution
For a complete solution, how do we know there exists a possible solution for n = 2 ? If there is, how do we know there is only one solution for n = 2 ? Fermat's Last Theorem states that there exists no solutions to the equation x n + y n = z n for n > 2 . It does not guarantee the existence of a solution for n ≤ 2 . Thus, you need to show that there is a solution for n = 2 , and that there is one , unique solution.
There are absolutely at least two solutions of the ordered pair (a,b) when n=2. I've found that (5,2) and (313,24) are distinct solutions. All that is required is a Pythagorean triple with its hypotenuse equal to one more than another one of its sides.
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How did you find these?
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I looked up a list of Pythagorean triples that satisfied the condition. It would be wonderful to see a proof that allows infinite, or limits the amount of solutions when n=2.
Would help if it was clarified whether 0 ∈ N or 0 ∈ / N . In France apparently it is common for 0 to be natural, which would make the answer 2 :)
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You're absolutely right! Thank you, I'll change it now.
Let x = b + 1 and y = a − 1 .
a n = x n + y n
Now we can use Fermat's Last Theorem to see that n < = 2 .
n = 1 , 2 fit the theorem and the definition as a natural number.
Although consider that when n = 1 , then a = b + 1 + a − 1 .
Rearrange:
a − a = b + 1 − 1
0 = b
As b = 0 (a number not part of the natural numbers), a contradiction occurs when n = 1 .
Therefore there is only one solution for n when n = 2 .