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#1 h Apr 26, 2019, 2:10 PM • 1 Y

Y by Adventure10

Find all triple ( x, y, z ) of natural numbers satisfying the equation $1+4^x+4^y=z^2$ .

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#2 h May 11, 2023, 8:58 AM

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It can be written as $4^x+4^y=(z-1)(z+1)$
Let's assume that $x>=y$ and let $k=x-y$
$4^y(4^k+1)=(z-1)(z+1)$ and because
$(4^y,4^x+1)=1$ we have 2 options:
$z-1=4^y$ and $z+1=4^k+1$ => $4^y+1=4^k$ and has no solutions or $z-1=4^k+1$ and $z+1=4^y$ => $4^k+1=4^y$
This has the only solution $(x,y,z)=(1,1,3)$

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#3 h Apr 10, 2025, 5:35 PM

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We start off by supposing W.L.O.G that x>=y. Suppose that 2y-1 < x, then 4^y + 4^x + 1 = 2 * 2^(2y-1) + 2^2x + 1 < 2 * 2^x + 2^2x + 1 = (2^x + 1)^2. But, 4^x + 4^y + 1 > 4^x = (2^x)^2 which means that (2^x)^2 < z^2 < (2^x + 1)^2 which in natural numbers is not possible, contradiction, then 2y-1 >= x. Now, 4^x + 4^y = 4^y(4^(x-y) + 1) = (z-1)(z+1) which means that z-1, z+1 are divisible by 2. v2(gcd(z-1, z+1)) = 1. First case: v2 (z-1) = 2y-1, v2(z+1) = 1. z-1 = 2^(2y-1) * k => 4^x + 4^y + 1 = z^2 = (2^(2y-1) * k)^2 = 4^(2y-1) * k^2 + 4^y * k + 1 >= 4^x * k^2 + 4^y * k + 1 => k=1 and 2y-1 = x. From this we get a solution (2y-1, y, 2^(2y-1) + 1) for all natural y. Second case: v2(z-1) = 1, v2(z+1) = 2y-1. z+1 = 2^(2y-1) *k => 4^x + 4^y + 1 = z^2 = (2^(2y-1) * k - 1) ^2 >= 4^x * k^2 - 4^y * k + 1 => 4^x * (k-1)(k+1) <= 4^y * (k+1). If k≠1, then 4^x <= 4^x * (k-1) <= 4^y => x=y, k=2 and z^2 = 1+2*4^y=1 + 2^(2y+1) = (2^2y - 1)^2 = 4^2y - 2^(2y+1) + 1 => 4^(y+1) = 4^2y + 1 > 4^2y which is not possible if y is a natural number. Then let k=1 => z^2 = (2^(2y-1) - 1)^2 = 4^(2y-1) - 4^y + 1 => 4^x = 2^(4y-2) - 2^(2y+1). If y=1 then 4^x = 4 - 8 < 0, then y>1, we get 2^2x = 2^(2y+1) * (2^(2y-3) - 1). Then we see that 2x = 2y+1 which is not possible, contradiction. From this, the only solution is (2y-1, y, 2^(2y-1) + 1) for all natural y.

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