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Pythagorean Quadruple

Stage: 3 and 4 Short Challenge Level: Challenge Level:1

It is clear that each of a, b and c must be less than or equal to $10$. A brief inspection will show that the only combination of different square numbers which total $121$ is $81, 36, 4$, so the $a, b$ and $c$ are $2, 6$ and $9$.
 
 
 

This problem is taken from the UKMT Mathematical Challenges.
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