# Determine the lengths of the sides of a right triangle

The positive real numbers $a,b,c$ are such that $a^2+b^2=c^2$, $c=b^2/a$ and $b-a=1$. Determine $a,b,c.$

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Hint: plug in what you are given. Plug $b=a+1$ into the second, then the second into the first, and what do you get? – Ross Millikan Jul 3 '13 at 13:39
Yet another textbook artificial question. – whatever Jul 3 '13 at 21:27
To my knowledge this particular problem has never appeared in any textbook. – Kelly Henrehan Jul 3 '13 at 21:35

From $a^2+b^2=c^2$ , $c=b^2/a$ and $b-a=1$, $a^2+b^2 = b^4/a^2$, so $a^4+a^2b^2 = b^4$.

We could substitute $b = a+1$ (if we do, I get $a^4-2a^3-5a^2-4a-1=0$, which I would rather not try to solve), but instead we will multiply by $4$ and use $4x^2-4x+1 = (2x-1)^2$.

\begin{align} 4a^4 &= 4b^4-4a^2b^2\\ 4a^4+a^4 &= 4b^4-4a^2b^2+a^4\\ 5a^4&=(2b^2-a^2)^2 \end{align}

so, taking the square root with $2b^2-a^2$,

\begin{align} a^2\sqrt{5} &= 2b^2-a^2\\ a^2(\sqrt{5}+1) &= 2b^2\\ a\sqrt{\sqrt{5}+1} &= b\sqrt{2}\\ &= (a+1)\sqrt{2}\\ a(\sqrt{\sqrt{5}+1}- \sqrt{2}) &= \sqrt{2}\\ a &= \frac{\sqrt{2}}{\sqrt{\sqrt{5}+1}-\sqrt{2}}\\ b &=a+1\\ &= \frac{\sqrt{2}+\sqrt{\sqrt{5}+1}-\sqrt{2}}{\sqrt{\sqrt{5}+1}-\sqrt{2}}\\ &= \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}-\sqrt{2}}\\ \end{align}

To get $c$, let $u = \sqrt{\sqrt{5}+1}$ and $v = \sqrt{2}$, so $a = \frac{v}{u-v}$ and $b = \frac{u}{u-v}$.

The first equation for $c$ is $c = \frac{b^2}{a} = \frac{u^2}{(u-v)^2}\frac{u-v}{v} =\frac{u^2 }{v(u-v)}$ or $c^2 = \frac{u^4 }{v^2(u-v)^2}$.

The second equation for $c$ is $c^2=a^2+b^2 =\frac{v^2}{(u-v)^2} +\frac{u^2}{(u-v)^2} =\frac{u^2+v^2}{(u-v)^2}$.

For the two expressions for $c^2$ to be equal, we need $\frac{u^4 }{v^2(u-v)^2} =\frac{u^2+v^2}{(u-v)^2}$ or $v^2(u^2+v^2) = u^4$.

To check that these are equal:

\begin{align} u^2 &= \sqrt{5}+1\\ u^4 &= 6+2\sqrt{5}\\ v^2 &= 2\\ u^2+v^2 &= \sqrt{5}+3\\ v^2(u^2+v^2) &=2(\sqrt{5}+3)\\ &= 2\sqrt{5}+6\\ \end{align}

so the two are equal (whew!).

We see that $c = \frac{u^2 }{v(u-v)} = \frac{\sqrt{5}+1}{\sqrt{2}(\sqrt{\sqrt{5}+1}-\sqrt{2})} = \frac{\sqrt{5}+1}{\sqrt{2}\sqrt{\sqrt{5}+1}-2}$.

Note: If we take the square root with $a^2-2b^2$, we get

\begin{align} a^2\sqrt{5} &= a^2-2b^2\\ a^2(1-\sqrt{5}) &= 2b^2\\ \end{align}

which can not be since the left side is negative and the right side is positive.

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I noticed that also and fixed it. Thanks. – marty cohen Jul 3 '13 at 15:14

Use $b-a=1$ to find an expression for $b$ in terms of $a$ (very easy), then substitute this into $c=\frac{b^2}{a}$ to get $c$ in terms of $a$. now you can write $a^2 + b^2 = c^2$ in terms of $a$. Then solve the quadratic to find $a$ and you can get the other terms from the expressions you have already derived. I haven't worked out what the roots of $a$ will be, but only one of them will be positive and give positive values for $b$ and $c$.

You can use this approach whenever you have $n$ equations in $n$ unknowns, although sometimes it can be difficult to see where to start. Here, it is fairly obvious to start with the simplest equation and build up.

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You get a quartic for $a$, not a quadratic. – marty cohen Jul 3 '13 at 21:12
Quite right, didn't realize it would be that tricky. Like your solution. – Baron Mingus Jul 5 '13 at 11:07

$$a^2+b^2=\frac{b^4}{a^2} \Rightarrow a^4+a^2b^2=b^4 \Rightarrow \frac{a^4}{b^4}+\frac{a^2}{b^2}=1\,.$$

Let $x =\frac{a^2}{b^2}$ then, as $x \geq 0$

$$x^2+x=1 \Rightarrow x=\frac{-1+\sqrt{5}}{2} \,.$$

Thus $$\frac{a}{b}= \sqrt{\frac{-1+\sqrt{5}}{2}}$$

This implies that

$$1- \frac{1}{b}=\frac{b-1}{b}= \sqrt{\frac{-1+\sqrt{5}}{2}} \Rightarrow b =\frac{1}{1-\sqrt{\frac{-1+\sqrt{5}}{2}}} \,.$$

Now all you need is to simplify these expressions....

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