Pre RMO 2016

Number Theory Level 3

Let m m and n n be non-negative integers satisfying 5 ⋅ 3 m + 4 = n 2 5 \cdot 3^m + 4 = n^2 , find the sum of all possible values of n n .

See my picks for RMO 2016


The answer is 10.

Md Zuhair by Md Zuhair , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Md Zuhair
Oct 6, 2016

In this question we have 5 ⋅ 3 m + 4 = n 2 5 \cdot 3^m + 4 = n^2 . Hence now 5 ⋅ 3 m = n 2 − 4 5 \cdot 3^m = n^2-4 . Hence 5 ⋅ 3 m = ( n − 2 ) ( n + 2 ) 5 \cdot 3^m = (n-2)(n+2) . Hence There are two cases .

Case 1 - n + 2 = 5 n+2=5

Case 2 - n − 2 = 5 n-2 = 5 .

Getting n from two equations as 3 3 and 7 7 respectevly.

Hence sum of all n's = 10 10

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Why do we have those two cases only?

Anik Mandal - 4 years, 8 months ago

Log in to reply

This two cases are required for n's value . See we will get m after getting the n. Then taking n for those cases of 3^m we get it .

Md Zuhair - 4 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...