Solve in $\mathbb{N}$ the system: $x²-y²=5440$ and $ x \land y = 8$

Question

We have:

$$x^{2}-y^{2} = (x-y)(x+y) = 5440 = 2^6\cdot 5\cdot 17$$

$$x \land y = 8.$$

I don't know how to proceed.

Thank you for your help.

Answer 1

Let $x=8a$ and $y=8b$, where $a \perp b$.

Then:

$$\begin{array}{rcl} (x-y)(x+y) &=& 2^6 \times 5 \times 17 \\ (8a-8b)(8a+8b) &=& 2^6 \times 5 \times 17 \\ (a-b)(a+b) &=& 5 \times 17 \\ \end{array}$$

Case one:

$$\begin{array}{cl} &\begin{cases}a-b&=&1\\a+b&=&85\end{cases} \\ \implies&\begin{cases}a&=&43\\b&=&42\end{cases} \\ \implies&\begin{cases}x&=&344\\y&=&336\end{cases} \end{array}$$

Case two:

$$\begin{array}{cl} &\begin{cases}a-b&=&5\\a+b&=&17\end{cases} \\ \implies&\begin{cases}a&=&11\\b&=&6\end{cases} \\ \implies&\begin{cases}x&=&88\\y&=&48\end{cases} \end{array}$$

Answer 2

As $\gcd (x,y)=8$ then you know $x =8x'$ and $y=8y'$ and $x^2-y^2=64 (x'^2-y'^2)=5440$

So $(x'^2-y'^2)=(x'-y')(x'+y')=85=5*17$

As $x'+y'$ is positive, then $x'-y'$ is positive so $x'>y'$ and $x'+y' > x'-y'$

The only ways to factor $85$ is either:

1) $x'+y'= 85;x'-y'=1$

Or

2) $x'+y'=17;x'-y'=5$

1) yields $y'=42;x'=43$ and $x=43*8=344;y=42*8=336 $

2) yields $y'=6;x'=11;x=88;y=48$

Answer 3

$$gcd(x,y) = 8$$

So $x = 8u$, $y = 8v$, so

$$(8u)^2 - (8v)^2 = 5440$$ $$u^2 - v^2 = 85$$ $$(u + v)(u - v) = 17 \times 5$$ or $$(u + v)(u - v) = 85 \times 1$$