LVII OM - I - Problem 1

Determine all non-negative integers $ n $ for which the number
$ 2^n +105 $ is a perfect square of an integer.
If $ n $ is an odd number, then the number $ 2^n +105 $ gives a remainder of 2 when divided by 3. Since a number that is the square of an integer cannot give a remainder of 2 when divided by 3, the number $ n $ satisfying the conditions of the problem must be even. 
Let's assume that $ n =2k $ for some non-negative integer $ k $. We want to solve the equation $ 2^{2k} +105 = m^2 $ in non-negative integers $ k $ and $ m $. We rewrite this equation as $ (m-2^k)(m+2^k) = 105 $. 
From the obtained equation, it follows that $ m-2^k >0 $. Of course, the inequality $ m+2^k > m-2^k $ is also satisfied. Since $ 105=3\cdot 5\cdot 7 $, finding non-negative integers $ m $ and $ k $ that satisfy the relation $ (m-2^k)(m+2^k)=105 $ reduces to solving four systems of equations:






By subtracting the first equation from the second, we conclude that the number $ 2 \cdot 2^k =2^{k+1} $ equals 8, 16, 32, or 104. From this, we obtain three possible values for $ k $; they are: 2, 3, or 4. Thus, the possible values of the number $ n $ are 4, 6, or 8. 
We directly check that for the three obtained values of $ n $, the number $ 2^n +105 $ is the square of an integer:











直接检查结果如下：
- 当 $ n = 4 $ 时, $ 2^4 + 105 = 16 + 105 = 121 = 11^2 $.
- 当 $ n = 6 $ 时, $ 2^6 + 105 = 64 + 105 = 169 = 13^2 $.
- 当 $ n = 8 $ 时, $ 2^8 + 105 = 256 + 105 = 361 = 19^2 $.

Directly checking the results:
- For $ n = 4 $, $ 2^4 + 105 = 16 + 105 = 121 = 11^2 $.
- For $ n = 6 $, $ 2^6 + 105 = 64 + 105 = 169 = 13^2 $.
- For $ n = 8 $, $ 2^8 + 105 = 256 + 105 = 361 = 19^2 $.