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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 $81+36+4$.
More formally, the problem can be analysed by considering the remainders after dividing the square numbers less than $121$ $(1,4,9,16,25,36,49,64,81$ and $100)$ by three. The remainders are $(1,1,0,1,1,0,1,1,0$ and $1)$. 
When $121$ is divided by $3$, the remainder is $1$. Therefore $a^2+b^2+c^2$ must also leave a remainder of $1$.

Now we can deduce that two of the three squares must leave a remainder of $0$ and so be multiples of $3$. There are three square numbers below $121$ which are multiples of three: $9, 36$ and $81$. Checking these, we see that $81$ and $36$ are the only pair to have a sum which differs from $121$ by a perfect square, namely $4$. 
So $a+b+c=9+6+2=17$.

This problem is taken from the UKMT Mathematical Challenges.
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