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ktreyb
7, 24, 25 makes one. 9, 40, 41 as well. 39, 52, 65 works. 11, 60, 61 does too. Lastly, 40, 42, 58 is a right triangle as well. 

How I found out was plugging each into the Pythagorean Theorem for proving/solving right triangles of a² + b² = c². The last number in each combo is your C. The other two numbers are interchangeable in regards to position. 
check the picture below.

let's do the red ones, to see if they check out.

[tex]\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}\\\\ -------------------------------\\\\ \sqrt{24^2+69^2}~~~~\approx~~~~ 73.054774~~~~\ne 74 \\\\\\ \sqrt{8^2+15^2}~~~~=~~~~17\ne 19 \\\\\\ \sqrt{30^2+60^2}~~~~\approx~~~~67.08204\ne 68[/tex]

and now, you can check any of the others, to see if the root of that sum is the last or longest side, namely the hypotenuse.
Ver imagen jdoe0001

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