Determine all possible integers n such that:

             abs(2n+5n-65)

is a perfect square.

Provide valid reasoning for your answer.

*** abs denotes the absolute value function.

Apparently, the only solutions to the function f(n) = abs(2^n+5^n-65) are:
f(2) = 6^2
f(4) = 24^2

Proof:
Negative values of n lead to non-integer values of the function between 64 and 65, and n=0 yields f(0) = 63.

For positive integers, the final digit of the function cycles between the nonzero even digits.  When n mod 4 is {0,1,2,3} the final digit is {6,2,4,8}.
So if n is odd, the function can never be a square since squares never end in 2 or 8.

For positive even n < 5, the solutions n=2 and n=4 have been found by direct calculation.

For n=6, the square root of f(6) is slightly less than 125 or 5^3
For n=8, the square root of f(6) is slightly more than 625 or 5^4
For even n >= 8, the 5^n term dominates the function.  Since n is even, let n = 2k.  So the square root of the function (5^2k + 2^2k - 65) is always slightly more than 5^k.  How much more?

The next higher square after (5^k)^2 is (5^k + 1)^2.

Compare 5^2k + 2*5^k + 1  with 5^2k + 2^2k - 65.
2*5^k + 1 vs 2^2k - 65
2*5^k + 66 vs 4^k
5^k + 33 vs 4^(k-1)
5^k + 33 is always greater than 4^(k-1)

Posted by Larry on 2023-01-15 07:59:18