Cute NT ^-^

For students of class 6-8 (age 12 to 14)
Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm
Cute NT ^-^

Unread post by Asif Hossain » Sun Apr 25, 2021 9:43 am

Find all positive integers $x,a,b$ such that $x^a+x^b=x^{a+b}$
Solution:
rewrite it as $x^a=x^b(x^a-1)$ but since $x^a \perp x^a-1$ the only possiblity is $x^a-1=1$ yieldin the only solution $(x,a,b)=(2,1,1)$.
Hmm..Hammer...Treat everything as nail

Marzuq
Posts:9
Joined:Mon Apr 05, 2021 4:30 pm

Re: Cute NT ^-^

Unread post by Marzuq » Thu May 20, 2021 12:15 pm

$x^{a} × x^{b} = x^{a+b}$
If $x^{a} = m,x^{b} = n$. Then,
$m + n = mn$
$n = m(n-1)$
Then $n-1 = 1 \rightarrow n=2$
It follows that $m=2$

So only solution $(x,a,b)=(2,1,1)$

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