You can also consider the prime factorization: a^{5a}=b^b => (p1^a1 * ... * pn^an)^{5a}=(p1^b1 * ... * pn^bn)^b, i.e. ak*5a=bk*b for k=1,...,n. This can be written as bk/ak=5a/b. Now, if bk/ak=5a/b 1, i.e. a|b => b=n*a. Then bk/ak=5/n>1, so n=1,2,3,4. Finally (na)^{na)=a^{5a} => n^n*a^n = a^5 and checking n=1,2,3,4 gives solutions only for n=1 and n=4.
3:07 you don't need both A and B to be >1 for the left side to be really a fraction, you just need B>1 (since A and B are coprime). Therefore you only need the two cases B>1 and B=1 instead of your three cases, which is reflected by the answer (a, b) = (1, 1) appearing twice in your solution.
surprised how much forgotten about power rules. Thought noting that a,b are powers of the same root would have been a more efficient way of proving over the GCD.method. Boy was i wrong!!! however less cases to check. Looks like the most efficient proof is to recognize b=n*a where a,b,n are powers of same root
You can also consider the prime factorization: a^{5a}=b^b => (p1^a1 * ... * pn^an)^{5a}=(p1^b1 * ... * pn^bn)^b, i.e. ak*5a=bk*b for k=1,...,n. This can be written as bk/ak=5a/b. Now, if bk/ak=5a/b 1, i.e. a|b => b=n*a. Then bk/ak=5/n>1, so n=1,2,3,4. Finally (na)^{na)=a^{5a} => n^n*a^n = a^5 and checking n=1,2,3,4 gives solutions only for n=1 and n=4.
I'm sorry but I don't think thus notation is at all clear..
@@leif1075 What notation? Can you specify?
@@digxx I also did it in same way. it becomes lot more easier
3:07 you don't need both A and B to be >1 for the left side to be really a fraction, you just need B>1 (since A and B are coprime). Therefore you only need the two cases B>1 and B=1 instead of your three cases, which is reflected by the answer (a, b) = (1, 1) appearing twice in your solution.
Good solution
the audio volume is too low
Agree
🔉🦻
True
brilliant solution!🔥🔥🔥
surprised how much forgotten about power rules. Thought noting that a,b are powers of the same root would have been a more efficient way of proving over the GCD.method. Boy was i wrong!!! however less cases to check. Looks like the most efficient proof is to recognize b=n*a where a,b,n are powers of same root
Mahn excellent solution but the your voice sound is really low
Diophantine equations OMG