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The sum and the product of two numbers are each equal to $$s\; +\; \frac{1}{s}\; +\; 2$$. What is the difference between the squares of the reciprocals of the numbers?

I started by letting the smaller number be $x$ and the larger be $y$. From the problem statement, $$xy = x+y = s\; +\; \frac{1}{s}\; +\; 2 = \frac{\left( s+1 \right)^{2}}{s}$$

I want to find $$\frac{1}{x^{2}}-\frac{1}{y^{2}}\; =\; \frac{\left( y+x \right)\left( y-x \right)}{\left( xy \right)^{2}}$$

However, at this point I am stuck as I need to express both $x$ and $y$ in terms of $s$.

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  • $\begingroup$ Yeah, find that in terms of $s$ $\endgroup$ – 1110101001 Feb 18 '14 at 21:20
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From $y=s+1/s+2-x$ it follows $0=xy-s-1/s-2=- x^2s + x(s+1)^2 - (s+1)^2$, which means $x=s+1$ or $x=(s+1)/s$. It follows,

$$ \frac{1}{x^2}-\frac{1}{y^2}=\frac{s^2-1}{(s+1)^2}, or $$

$$ \frac{1}{x^2}-\frac{1}{y^2}=\frac{1-s^2}{(s+1)^2}. $$

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You are almost there with your expression

$\large \frac{1}{x^2}-\frac{1}{y^2}=\frac{(y+x)(y-x)}{(xy)^2}$

As we know that $(x+y)=xy$, this simplifies to

$\large \frac{1}{x^2}-\frac{1}{y^2}=\frac{(y-x)}{xy}$

The tricky part is to find $(y-x)$

Now if we square $(x+y)$ we obtain

$(x + y)^2 = x^2+y^2+2xy=(xy)^2$

so that

$x^2+y^2=(xy)^2-2xy=xy(xy-2)$

we can use this result as below.

$(y-x)^2=(x^2+y^2)-2xy=(xy(xy-2))-2xy=xy(xy-4)$

Thus

$\large (y-x)=\pm\sqrt{xy(xy-4)}=\pm\sqrt{\frac{(s+1)^2}{s}\frac{(s+1)^2-4s}{s}}=\pm\sqrt{\frac{(s+1)^2}{s}\frac{(s-1)^2}{s}}$

This simplifies to

$\large (y-x)=\pm \frac{(s^2-1)}{s}$

leading to

$\large \frac{1}{x^2}-\frac{1}{y^2}=\frac{(y-x)}{xy}=\pm \frac{(s^2-1)}{s}\frac{s}{(s+1)^2}=\pm\frac{(s^2-1)}{(s+1)^2}$

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