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Triangular Triples

Age 14 to 16
Challenge Level

Chong Ching Tong from River Valley High School, Singapore and Andrei Lazanu, age12, School: No. 205 Bucharest, Romania approached this problem in different ways.

Here is Chong's working:

\begin{eqnarray} 8778 \times 2 = 17556\\ \sqrt{17556} \cong 132.5\\ \frac {(132 \times 133)}{2} = 8778\\ 10296 \times 2 = 20592\\ \sqrt{20592} \cong 143.5\\ \frac {(143 \times 144)}{2} = 10296 \\ 13530 \times 2 = 27060\\ \sqrt{27060} \cong 164.5\\ \frac {(164 \times 165)}{2} = 13530\\ (8778)^2 + (10296)^2 = (13530)^2 \end{eqnarray}

Here is Andrei's solution:

First I demonstrate that 8778, 10296 and 13530 are triangular numbers, i.e. they can be written in the form n(n+1)/2. In order to do this I decomposed the product of each of the three numbers by 2 in the hope to put it in the form n(n+1)/2. I found: $$8778\times2 = 2^2\times3\times 7\times 11\times 19 = 132\times133$$ So, $$8778 = \frac{132 \times 133}{2}$$ and so 8778 is a triangular number. $$10296\times 2 = 2^2\times 11\times13 = 143\times144$$ So, $$0296 = \frac{143 \times 144}{2}$$, and it is a triangular number. $$13530\times 2 = 2^2\times3\times5\times11\times41 = 164\times 165$$ So, $$3530 = \frac{164 \times 165}{2}$$ is also a triangular number.

Now, I demonstrate that the three numbers are a Pythagorean triple. The greatest number is 13530 $$13530^2= 183060900$$ $$8778^2+ 10296^2 = 77053284 + 106007616 = 183060900$$

So, the three numbers are a Pythagorean triple