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# find all pairs of positive integers a,b such that a^b-b^a=3

4 years ago

mycroft holmes
272 Points

Since RHS is odd, we must have either a is even and b is odd or vice versa.

Case I: a is even, b is odd

Also if $b \ne 1$, we must haveclearly, b>a. Hence $a \ge 2$ and hence $b \ge 3$. (In fact looking modulo 3, a is an even number of the form 3k+1, and hence the least value of a is 4, but that is not required here)

Now $a^b$ will be of the form 8k, whereas $b^a$ being the square of an odd number will be of the form 8k+1. So in the equation$a^b = b^a+3$, LHS is of the form 8k while RHS is of the form 8k+4, which is a contradiction.

Case II: a is odd, b is even. Again, $a \ge 3$, So in the equation $a^b = b^a+ 3$, LHS is of the form 8k+1 while RHS is of the form 8k+3 and so no solutions are possible.

That only leaves us with the possibility, b =1, in which case a = 4. Hence (4,1) is the only solution in natural numbers.

4 years ago
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