I'm having difficulty remembering wot to do with an equation lyk this, im pretty sure its slightly different to the last one i posted and i just cant remember or find it in my textbooks. Any help would b fantastic.
3 x (2^x) + 1 = n^2
You must find all pairs of positive integers that work for this equation.
Cant remember wot method 2 use.
Thats wot im having problems with. Is it lyk an infinite number of pairs??
so would i attain 3 x (2^m) + 1 = n^2
Would i substitute n's value with terms of m, or would i just say theres an infinite number of pairs?
Im still working on it, cant think wot it is, ideas??
First case:
so,
But on the left side we have also , hence must have a factor of , but when this happens? It happens when have a factor of :
hence, have a factor of , but for which ?
It easy to show that must be odd.
Now, do the same for the second case, but now when .
At the and you will get a collection of infinite solutions.
That is not true. n=5 , x=3 shows . but neither n-1, nor n+1 is 8.
pikachu26134,
Here are some exercises for you,
1) Show that if n > 2 satisfies your equation,n must be odd.
1') Substitute n=2 and see if it satisfies your equation.
2) Show that if and only if OR for some integer k such that
3) By subtracting the equations in the above system individually, conclude that k=1 is the only possible scenario. Thus arrive at the conclusion that only two solutions with odd 'n' are possible.
So totally there are 3 solutions. Show me some effort by trying to solve the above simpler exercises and I will help you further.