Pythagorean Triple and Powers of Two

Since $a$ is a power of $2$ and $8$ is also a power of $2$ , $8a$ is also a power of $2$ and $\implies b = 8a-1$ and $c = 8a+1$ . Since $c$ is the largest, we have:

$\begin{aligned} a^2 + b^2 & = c^2 \\ a^2 + (8a-1)^2 & = (8a+1)^2 \\ a^2 + 64a^2 - 16a + 1 & = 64a^2 + 16a + 1 \\ a^2 & = 32a \\ \implies a & = 32 & \small \blue{\text{a power of 2}}\end{aligned}$

Therefore $a+b+c = 32 + 8\times32 - 1 + 8 \times 32 + 1 = 17 \times 32 = \boxed{544}$ .

Let the integers of the triplet be $x^2-y^2, 2xy, x^2+y^2$ . Let $2xy=2^n\implies xy=2^{n-1}, x^2-y^2=2^p-1, x^2+y^2=2^p+1$ . Then $x=2^{p/2}, y=1$ . Therefore $p=2(n-1)$ . Also, $8\times 2xy=2^p\implies 2^{n+3}=2^{2(n-1)}\implies n=5$ . Therefore $a=2xy=2^5=32, b=2^8-1=255, c=2^8+1=257$ and $a+b+c=32+255+257=\boxed {544}$ .

By using the first and the last point, we get two equations.

$a^2 + b^2 = c^2$

$8a = c - 1$

Now let's use the remaining points and let

$\textcolor{#20A900}{a} = 2^\alpha$

$\textcolor{#20A900}{b} = 2^\beta - 1$

$\textcolor{#20A900}{c} = 2^\beta + 1$

for some integers $\alpha, \beta \in \mathbb{N}$ .

By substituting for $\textcolor{#20A900}{a}, \textcolor{#20A900}{b}, \textcolor{#20A900}{c}$ , we get

$a^2 + b^2 = c^2$

$(2^\alpha)^2 + (2^\beta - 1)^2 = (2^\beta + 1)^2$

Mind that $a^m \cdot a^n = a^{m + n}$ and $(a^m)^n = a^{m \cdot n}$ .

$2^{2\alpha} + 2^{2\beta} - 2 \cdot 2^\beta + 1 = 2^{2\beta} + 2 \cdot 2^\beta + 1$

$2^{2\alpha} = 4 \cdot 2^\beta$

$2^{2\alpha} = 2^2 \cdot 2^\beta$

$2^{2\alpha} = 2^{2 + \beta}$ .

By eliminating the basis:

$\textcolor{#3D99F6}{2\alpha = 2 + \beta}$ .

Now substitute $\textcolor{#20A900}{a}, \textcolor{#20A900}{b}, \textcolor{#20A900}{c}$ into the second equation.

$8a = c - 1$

$8 \cdot 2^\alpha = (2^\beta + 1) - 1$

$2^3 \cdot 2^\alpha = 2^\beta$

$2^{3 + \alpha} = 2^\beta$

$\textcolor{#3D99F6}{3 + \alpha = \beta}$ .

We have obtained a system of two equations with two variables which we can solve easily.

$\textit{(I)} \thinspace 2\alpha = 2 + \beta$

$\underline{\textit{(II)} \thinspace 3 + \alpha = \beta}$

$\textit{(I)} - \textit{(II)}:$

$2\alpha - \alpha - 3 = 2 + \beta - \beta$

$\alpha - 3 = 2$

$\alpha = 5$

By substitution into $\textit{(II)}$ :

$3 + 5 = \beta$

$\beta = 8$ .

Finally, we just subtitute $\alpha, \beta$ back into $a, b, c$ and sum them up.

$\textcolor{#20A900}{a} = 2^\alpha = 2^5 = 32$

$\textcolor{#20A900}{b} = 2^\beta - 1 = 2^8 - 1 = 255$

$\textcolor{#20A900}{c} = 2^\beta + 1 = 2^8 + 1 = 257$

$a + b + c = 32 + 255 + 257 = \textcolor{#D61F06}{544}$

We can confirm that $32^2 + 255^2 = 257^2$ .

Let x = largest ( 1 more than a square)

x - 2 = middle ( 1 less than a square)

(x-1)/8 = smallest

x^2 =(x-2)^2+((x-1)/8)^2

x^2=x^2-4x+4+(x^2-2x+1)/64

64x^2=64x^2-256x+256+x^2-2x+1

0=x^2-258x+257

0= (x-257)*(x-1)

x=257 or x=1(not 1 more than square)

x-2=255

(x-1)/8=32

Sum = 544