y ≡ x - 16
to allow factoring the term 2¹⁶ completely.
Then 2¹⁶ + 2¹⁹ + 2ˣ
= 2¹⁶ + 2¹⁹ + 2ʸ⁺¹⁶
= 2¹⁶ . (1 + 2³ + 2ʸ)
= 2¹⁶ . (9 + 2ʸ)
Now a solution, viz y=4, is readily obvious as yielding
= 2¹⁶ . (9+16)
= 2¹⁶ . 25
= (2⁸ . 5)²
∴ x = y + 16 = 20.
Suppose
∃ a ∈ Z
such that
9 + 2ʸ ≡ a²
Then
2ʸ = a² - 9 = (a-3) . (a+3)
and both (a-3) and (a+3) must be powers of 2.
This only occurs for
a = 5 => 2ʸ = 16 => y = 4 => x = 20.
Our solution is unique.
(2⁸ . 5)²
= (2⁷ . 2 . 5)²
= (2⁷ . 10)²
= (128 . 10)²
= 1280²
= 2¹⁶ . 25
= 2¹⁴ . 4 . 25
= 2¹⁴ . 100
= 16,384 . 100
= 1,638,400.
2ᵗ - 2ˢ = 2ˢ . (2ᵗ⁻ˢ - 1)
Hence when
2ˢ . (2ᵗ⁻ˢ - 1) = 6 = 2 . 3
it is necessary that:
2ˢ = 2¹ = 2
2ᵗ = 2ˢ . (3+1) = 2 . 4 = 8.
Hence
a = 2ˢ + 3 = 5 = 2ᵗ - 3
as above.