An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
What is the smallest number with exactly 14 divisors?
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
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 $4$.
This problem is taken from the UKMT Mathematical Challenges.