Find the sum of all positive integers n for which |2^n+5^n-65 | is a perfect square.

Step-by-step solution

|2^n+5^n-65 | has unit digit same as 2^n i.e., 2,4,8,6, for n=1,2,3,4, Also for perfect square it shall be 4 or 6 Hence n can be even only for n=2 |2^2+5^2-65 |=36n=4 |2^4+5^4-65 |=576 Also (K^2+1 )^2-K^2=2 K+1 or K^2+(2 K+1)=(K+1)^2 Let | 5^n K^2 + 2^n-65 2 K+1 | to be (K+1)^2 i.e., for n=6K= (5^3 ) 2 K+1=251 but 2^6-65=-1 for n=8K=5^4 2 K+1=626 but 2^8-65 191 for n=10K=5^5 2 K+1=3126 but 2¹⁰-65 959 The difference increases as 2 (5^x )+1 rises faster than 2^x-65 Hence only n=2 and 4 are possible