Questions   
Sort: 
 #2
avatar
+1
Nov 18, 2021
Nov 17, 2021
 #4
avatar+118654 
+1

Fist thing I did was graph it.

 

https://www.geogebra.org/classic/svfrstbf

 

a^2+b^2=ab^2

its symetrical so I want to look at the top half   b>0 (then I will double the number of answers)

 

\(a^2=ab^2-b^2\\ a^2=b^2(a-1)\\ b=\frac{a}{\sqrt{a-1}} \)

 

a has to be greater than 1,  

 

For b to be an integer  

a^2 = k (a-1)     where k is an integer

---------------------------------

yea I don't know.

There are at least 2 answers.

 

Nov 17, 2021
 #3
avatar
0
Nov 17, 2021

3 Online Users

avatar
avatar