# Theorem of Pythagoras - Incorrect Derivation

I am trying to solve below question from Coursera Intro to Calculus (link)

A right-angled triangle has shorter side lengths exactly $$a^2-b^2$$ and $$2ab$$ units respectively, where $$a$$ and $$b$$ are positive real numbers such that $$a$$ is greater than $$b$$. Find an exact expression for the length of the hypotenuse (in appropriate units).

Below are the choices

$$(a - b)^2$$

$$\sqrt(a^4 + 4a^2b^2 -b^4)$$

$$a^2 + b^2$$

$$\sqrt(a^2 + 2ab -b^2)$$

$$(a + b)^2$$

When I attempt to work out the solution (and I even got a 2nd pair of eyes to look at it, but he arrived at the same conclusion), I get this:

$$(a^2 - b^2)^2 + (2ab)^2 = x^2$$

$$a^4 -2a^2b^2 + b^4 + (2ab)^2 = x^2$$

$$a^4 -2a^2b^2 + b^4 + 4a^2b^2 = x^2$$

$$a^4 + 2a^2b^2 + b^4 = x^2$$

$$\sqrt(a^4 + 2a^2b^2 + b^4) = x$$

You are done. just multiply out $$(a^2 + b^2)^2$$ and check that you get $$x^2$$.

$$a^4+2a^2b^2+b^4 = (a^2+b^2)^2$$

Let $$a' = a^{2} - b^{2}$$ $$b' = 2ab$$

From the Pythagorean Theorem, $$c'^{2} = a'^{2} + b'^{2}$$

Substitute both $$a'$$ and $$b'$$ into the above equation:

$$c'^{2} = (a^{2} - b^{2})^{2} + (2ab)^{2} = a^{4} - 2a^{2}b^{2} + b^{4} + 4a^{2}b^{2} = a^{4} + 2a^{2}b^{2} + b^{4}$$

Solve for $$c'$$ by taking the square root:

$$c' = \sqrt{a^{4} + 2a^{2}b^{2} + b^{4}}$$

Here, $$a^{4} + 2a^{2}b^{2} + b^{4}$$ is a perfect square, and $$\sqrt{a^{4} + 2a^{2}b^{2} + b^{4}} = \sqrt{(a^2 + b^2)^{2}} = a^2 + b^2$$

• Is the minus sign on the last term of Option 2 not relevant? – DJohnM May 30 at 16:49
• Thank you for the correction. I will edit my answer accordingly. – Marvin May 30 at 16:52

$$c^2=(a^2-b^2)^2+(2ab)^2$$ $$=(a^4-2a^2b^2+b^4)+(4a^2b^2)$$ $$=a^4+2a^2b^2+b^4=(a^2+b^2)^2$$ $$\Rightarrow c=a^2+b^2$$