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Generating Triples

Stage: 4 Challenge Level: Challenge Level:1

Elliott from Wilson's school used a spreadsheet to find sets of triples where the hypotenuse is 1, 2, 3 and 4 units bigger than one of the other sides. He explains his method:

If you square each number, subtract one square from the square greater than it, then square root this number, you can find Pythagorean Triples. If the result is a whole number, the two numbers and the square rooted number make up a Pythagorean Triple.

For example, $24^2 = 576$, and $25^2 = 625$. $625-576 = 49 = 7^2$, so $(7, 24, 25)$ make up a Pythagorean Triple. 

The first square, when the hypotenuse is one greater than the other side, is always odd, increasing by two each time.

$9^2 + 40^2 = 41^2$
$11^2 + 60^2 = 61^2$
$13^2 + 84^2 = 85^2$

If, instead of subtracting the square from the square one greater than it, you subtract it from the square two greater, you can work out the Pythagorean Triples similar to Alison's.

For example, $15^2 = 225$ and $17^2 = 289$. $289 - 225 = 64 = 8^2$ so $(8, 15, 17)$ makes up a Pythagorean Triple.

The first square number is always even and increases by two each time.

$10^2 + 24^2 = 26^2$
$12^2 + 35^2 = 37^2$
$14^2 + 48^2 = 50^2$
 

If the hypotenuse is three longer than a side, the shortest side increases by 6 each time (starting on 3) e.g. 3, 15, 21, 27 etc. If the hypotenuse is four units longer, the shortest side increases by 4 each time e.g. 4, 8, 12, 16 etc.

Michael from St Patrick's in Australia used the following method:

All of Charlie's triples have an even square in them. Using that, the triples for Charlie are 
$1^2 + 0^2 = 1^2$
$3^2 + 4^2 = 5^2$
$5^2 + 12^2 = 13^2$
$7^2 + 24^2 = 25^2$
$9^2 + 40^2 = 41^2$
$11^2 + 60^2 = 61^2$
$13^2 + 84^2 = 85^2$
There is a pattern. The smaller, odd, square number will always rise by two. The larger, even, square number will go up by multiples of four in order, 4,8,12,16,20,24 etc. The answer will always be one square number more than the largest number in the equation.
The next equation in the pattern will be $15^2+112^2=113^2$. $15^2=225, 112^2=12544,$ and $113^2=12769$. $12544+225=12769$.

Thomas, also from St Patrick's sent us his work on Alison's Pythagorean triples:

The hypotenuse is two units longer than one of the other sides. Can you find a formula for generating Pythagorean Triples like Alison's? Can you prove that your formula works?

The three examples given are: 4²+3²=5² 6²+8²=10² 8²+15²=17² There is a pattern here (from left to right) the first number goes up by 2. The second number goes up by 5 then by 7 so it only makes sense that in the next equation the second number would be (15+9)=24. The third number again goes up by 5 then 7 so the third number should be (17+9) =26 Using this info I can make the accurate assumption that the next equation in line will be; 10²+24²=26². Now to prove this, 10²=100 and 24²=576 so 100+576=676. 26²=676. The next two equations in line will be 12²+35²=37² 14²+48²=50²

I wonder if you can explain why Michael's and Thomas's patterns generate Pythagorean triples?

Nathan from Rushmore School observed:

If $x+x+1=y^2$, $y,x,x+1$ make a Pythagorean triple.

Get two consecutive whole numbers that add up to a square number. For example, $4$ and $5$ - $4+5=9$ and $\sqrt9=3$ 3,4 and 5 make a Pythagorean triple. $12$ and $13$ - $12+13=25$ $\sqrt{25}=5$ 5,12 and 13 make a Pythagorean triple.
The inverse of this is get any odd number ($x$) and square it. Halve the square number and add/take away half to make $y$ and $z$. $x,y$ and $z$ will be Pythagorean triples.

Once you have these numbers you can multiply the triples by $n$ to make more Pythagorean triples.

Lucy from Belgium sent us her solution:

The formula is (2n+1)² + (2n²+2n)² = (2n²+2n+1)²

I tried it a few times :
If n=1 => 3² + 4² = 5²
If n=5 => 11² + 60² = 61²

Finding that you need an odd square number as the difference between the two other numbers wasn't so difficult. And then I found that between 4 and 12 there were 8 units, between 12 and 24 there are 12 units, and then 16 units and 20 units and so on... Which meant that the n was multiplied each time with the previous number+2
n=1 was multiplied by 4 to get to 4
n=2 was multiplied by 6 to get to
12 n=3 was multiplied by 8 to get to 24 and so on again... in other words n was multiplied by the odd square number plus one => n(2n+2) => (2n²+2n) the last number is just one unit more so that is (2n²+2n+1)

Well done Lucy! Can you show algebraically that the expression will always give a Pythagorean triple?

We also received solutions from Niharika, Chuyi and Rajeev.