# Find the number of four tuples (a,b,c,d) of positive integers satisfying all three equations a^3=b^2, c^3=d^2, c-a=64 ?

Jul 12, 2016

$\left\{a = {15}^{2} , b = {15}^{3} , c = {17}^{2} , d = {17}^{3}\right\}$ and
$\left\{a = {6}^{2} , b = {6}^{3} , c = {10}^{2} , d = {10}^{3}\right\}$

#### Explanation:

Solving

{(a^3=b^2), (c^3=d^2), (c-a=64) :}

for $b , c , d$ we obtain the feasible (integer positive) solution

$b = {a}^{\frac{3}{2}} , c = 64 + a , d = {\left(64 + a\right)}^{\frac{3}{2}}$

Regarding the solution for $b$ then $a = {m}^{2}$ and
regarding the solution for $d$ then $64 + a = {n}^{2}$

so

${n}^{2} - {m}^{2} = 64 = {2}^{6}$

or

$\left(n + m\right) \left(n - m\right) = p \cdot q = {2}^{6}$

and

$n + m = p = \left\{{2}^{6} , {2}^{5} , {2}^{4}\right\}$

For system

{ (n+m=p), (n-m=q) :}

the feasible solutions are

$\left\{n = 17 , m = 15\right\}$ and $\left\{n = 10 , m = 6\right\}$

then $a = \left\{{15}^{2} , {6}^{2}\right\}$
$d = \left\{{\left(\sqrt{64 + {15}^{2}}\right)}^{3} = {17}^{3} , {\left(\sqrt{64 + {6}^{2}}\right)}^{3} = {10}^{3}\right\}$
$c = \left\{64 + {15}^{2} = {17}^{2} , 64 + {6}^{2} = {10}^{2}\right\}$
$b = \left\{{15}^{3} , {6}^{3}\right\}$

Jul 13, 2016

There are two, namely:

$\textcolor{b l u e}{\text{(15^2, 15^3, 17^2, 17^3)) = color(blue)(} \left(225 , 3375 , 289 , 4913\right)}$

$\textcolor{b l u e}{\text{(6^2, 6^3, 10^2, 10^3)) = color(blue)(} \left(36 , 216 , 100 , 1000\right)}$

#### Explanation:

Let:

$p = \frac{b}{a}$

Then:

${p}^{2} = {b}^{2} / {a}^{2} = {a}^{3} / {a}^{2} = a$

${p}^{3} = {b}^{3} / {a}^{3} = {b}^{3} / {b}^{2} = b$

Then since $p$ is a positive rational number whose square is a positive integer, $p$ must be a positive integer.

Similarly, if we let $q = \frac{d}{c}$, then $q$ is a positive integer with:

${q}^{2} = c$

${q}^{3} = d$

Then:

${2}^{6} = 64 = c - a = {q}^{2} - {p}^{2} = \left(q - p\right) \left(q + p\right)$

Both $\left(q - p\right)$ and $\left(q + p\right)$ are positive integers and $\left(q - p\right) < \left(q + p\right)$.

So the only possible factorings of ${2}^{6}$ give us:

$\left\{\begin{matrix}\left(q - p\right) = 1 \\ \left(q + p\right) = 64\end{matrix}\right. \textcolor{w h i t e}{X X}$ hence $2 q = 65 , \textcolor{w h i t e}{X} \textcolor{red}{\cancel{\textcolor{b l a c k}{q = \frac{65}{2}}}}$

$\left\{\begin{matrix}\left(q - p\right) = 2 \\ \left(q + p\right) = 32\end{matrix}\right. \textcolor{w h i t e}{X X}$ hence $2 q = 34 , \textcolor{w h i t e}{X} \textcolor{b l u e}{q = 17} , \textcolor{w h i t e}{X} \textcolor{b l u e}{p = 15}$

$\left\{\begin{matrix}\left(q - p\right) = 4 \\ \left(q + p\right) = 16\end{matrix}\right. \textcolor{w h i t e}{X X}$ hence $2 q = 20 , \textcolor{w h i t e}{X} \textcolor{b l u e}{q = 10} , \textcolor{w h i t e}{X} \textcolor{b l u e}{p = 6}$

So there are two possible tuples $\left(a , b , c , d\right)$ satisfying the conditions, namely:

$\textcolor{b l u e}{\text{(15^2, 15^3, 17^2, 17^3)) = color(blue)(} \left(225 , 3375 , 289 , 4913\right)}$

$\textcolor{b l u e}{\text{(6^2, 6^3, 10^2, 10^3)) = color(blue)(} \left(36 , 216 , 100 , 1000\right)}$