If the sides of a right angled triangle are $a$, $b$ and $c$ units, with $c$ being the hypotenuse or longest side, we know by Pythagoras' Theorem that $a^2+b^2=c^2$.

Charlie has been looking for examples where $a$, $b$ and $c$ are all whole numbers. These sets are called *Pythagorean Triples*.

He drew a table:

From his table, he spotted two sets of Pythagorean triples:

$3^2+4^2=5^2$

$5^2+12^2=13^2$

**Can you find more Pythagorean Triples like Charlie's?**

Can you find a formula to generate them?

Can you prove that your formula works?

Alison has been working on Pythagorean Triples where the hypotenuse is 2 units longer than one of the other sides.

So far, she has found these:

$4^2 + 3^2 = 5^2$

$6^2+8^2=10^2$

$8^2+15^2=17^2$

Some of these are just scaled-up versions of Charlie's triples, but some of them are new and can't be divided by a common factor (these are called primitive triples).

**Can you find more Pythagorean Triples like Alison's?**

Can you find a formula for generating Pythagorean Triples like Alison's?

Can you prove that your formula works?

**Here are some other questions you might like to consider:**

- Can you find triples where the hypotenuse is 3 units longer than one of the other sides? Or 4 units longer? Or...?
- Can you say anything about when such triples will be primitive triples?
- Are there any triples where the two
*shorter*sides differ by one unit?

Share your thoughts and discoveries

## Comments

### square numbers

I noticed that if $a^2 + b^2 = (b+2)^2$, a and b have nothing in common but the hypotenuse is always two more than one of the other sides.

Which simplified gives us ... a = 2 times the square root of b+1.

To find b you just need to take away 1 from a square number, and to get the value of a you input b into the formula above.

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