See for the problem here: Click here
Correct Answer:
The above answer was guessed correctly by Abhishek M N (Teacher at Pragnadeepa), Ramya C S (Teacher at Arivu Vidya Samsthe), Gagan Dev (V Semester PCM, Sarada Vilas College), Vinya Kumar (V Semester PMCs, Sarada Vilas College), Manjunatha M R (Research Scholar, IIT Indore) and Nagendra P (Assistant Professor, GFGC Paduvalahippe, Hassan).
However the nearly convincing justifications through number theory were given by Ramya C. S. and Manjunatha M. R. For solution through calculus (which was not expected here), Click here.
Note. If you find any mistake, please write to us at dep.math.svc@gmail.com
Solution(s)
We first propose the genral steps in proofs of both Manjunatha M. R. and Ramya C. S.
- Step 1: Assuming \(a \lt b \), first show that \(a\) divides \(b\) (Manjunatha M. R. showed that \(a\) divides \(b\), while Ramya C. S. assumed this)
- Step 2: If \(a\) divides \(b\), then \(b = ka\) for some integer \(k\). Thus \(a^b = b^a\) becomes \[a^{ka} = (ka)^a. \] This shows that \[ a^k = ka\] Or \[ k = a^{k-1}.\]
- Step 3: Show that \(k = a^{k-1}\) is impossible to happen if \(a>3\) and \(k\geq 2 \) (This was not shown by both although they argued that this would be true.) This is indeed true. And hence, if \(k=a^{k-1}\), then either \(k=1\) in which case \(a=b\) which is not allowed or \(a\leq 2\). As \(a\) cannot be 1 (else, \(b\) would also be 1), \(a = 2\). In this case, show that only for \(k=2\), it is true that \(k = 2^{k-1}\).
- Step 4: Conclude that the only allowed solution is \(a=2\) and \(b = a^2 = 4\).
For the solution of Manjunatha M. R., Click Here
We present below how we could have justified Step 1 and Step 3 in a different way
Justification to Step 1: Assume that \(a \lt b\). We have \(a^b = b^a\). This means \(a^{b-a}a^a = b^a\). This works well in this context, as \(a\lt b\). Thus, this shows that \(a^a\) divdes \(b^a\). This implies \(a\) divides \(b\). This is a general fact in number theory that if \(a^n\) divdies \(b^n\), then \(a\) divides \(b\). Of course, this can be be seen very easily from fundamental theorem of artimetic that if for some prime \(p\), \(p^k\) divides \(a\), and \(p^{kn}\) divides \(b^n\), then \(p^k\) divides \(b\).
Justification to Step 3: This can be proved easily by mathematical induction. We can easily show that for \(a\geq 3\) and \(k\geq 2\), we have \(a^{k-1} \> k\). We shall show it just for \(a=3\) and for \(a \gt 3\), the result follows immediately. Note that, if \(a=3\), then for \(k=2\), we have \[a^{k-1} = 3^{2-1} = 3 > 2 = k.\] Thus, for \(k=1\), the result holds. Suppose, the result holds for some \(k=n\). That is, \[3^{n-1} > n.\] Multiply both the sides by \(3\) to get \[3^n = 3n.\] Clearly \(3n > n+1\) for any \(n\) as \(2n >1\) for any \(n\). Thus, it is also true for \(k=n+1\) that \[3^{k-1} > k .\] We leave it to the readers to prove in the same way for \[2^{k-1} > k \] for any \(k\geq 3\).
For a beautiful hirtorical accecdote, see the below image from wikipedia. Thanks to Manjunatha M. R. for pointing at it
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