Interesting indices
How many positive integer solutions are there to the following equation: a b = b a ?
where a = b and (a,b) is not considered to be different to (b,a)?
The answer is 1.
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2 solutions
First lets note that 2 4 = 4 2 . Now I am left to prove whether or not there are more solutions. Lets assume (without loss of generality) that a > b . This means that b a > a b , if both sides are more than 16 because if we keep b constant, then the left - hand side is an exponential function and the right side is of the form x n where x is a constant and n is the variable (i.e. 2 1 0 > 1 0 2 ). This means that there are no solutions above 16 because the exponential graph will have a far steeper gradient. Also, by testing values of a and b, such that both sides are less than 16, shows that there are no solutions below 16. Hence, 2 4 = 4 2 is a unique solution.
Is ( 4 , 2 ) , ( − 2 , − 4 ) , ( − 4 , − 2 ) are solutions of the given equation?
There are a lot of non-integer solutions ...
Are you sure that a > b ⇒ b a > a b ? For example, take a = 3 , b = 1 .
FYI: It is generally better to place the entire equation into 1 latex expression, instead of splitting them up separately.
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yes but I added the fact that both sides had to be more than 16 for a > b ⇒ b a > a b (if we only consider positive integers).
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Well, that statement still has to be proven.
If we allowed for real numbers, then taking a = 5 , b = 1 . 7 5 , we have b a = 1 6 . 4 1 3 < a b = 1 6 . 7 1 8 . As such, your claim is not immediately obvious. What is so special about integers?
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@Calvin Lin – That is a valid point, so I'll try to prove the condition here.... Firstly neither one of a or b can equal 1 otherwise we would have a = b = 1, which doesn't satisfy the conditions. Next we have 2 x , which is more than 16 if a ≥ 5. Above I have already proven that the sides would not be equal in this case (i.e. 2 5 > 5 2 , 2 6 > 6 2 etc...). For base 3: the exponent has to be more than or equal to 4, so we have 3 4 > 4 3 . Now we have to prove that n ( n + 1 ) > ( n + 1 ) n , by dividing by n n to give: n > ( 1 + 1 / n ) n and as n → ∞ the R H S → e , so the inequality is true. (as the gap between a and b gets larger, so does the gap between a b and b a as argued initially)
a^b=b^a implies ln(a)/a=ln(b)/b. By drawing the graph of ln(x)/x for positive x we see the maximum of this function occurs at x=e, y=1/e, with x-intercept at x=1. Now for the solutions to occur two different integer values of x must give the same y value. As the graph decreases from x=e to large x we see that the two integers must fall into two seperate groups. One integer between 1 and e, the other greater than e. Hence one of the integers has to be two, solving for the other value we find the solution (2,4). As there are no more possible lower integer values than 2 this is the only solution where a is not equal to b.