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Answer: 9 different remainders (0, 1, 2, 3, 4, 5, 6, 7, 16)

$n$ $n^2$ $n+4$ remainder
1 1 5 1
2 4 6 4
3 9 7 2
4 16 8 0
5 25 9 7
6 36 10 6
7 49 11 5
...     ...
Not much pattern so far
...     ...
20 400 24 24$\times$15 = 240 + 120 = 360
24$\times$16 = 360 + 24 = 384
remainder is 16
100 10000 104 104$\times$100 = 10400
104$\times$96 = 10400 - 4016
               = 10000 - 16
remainder is 16
101 10201 105 105$\times$100 = 10500
105$\times$97 = 10500 - 315
              = 10200 - 16
remainder is 16

Remainder is 16 for the larger numbers tested
Had to multiply by $n-4$ for the larger values of $n$, try this algebraically: $(n-4)(n+4)=n^2-16$

So $n^2  = \underbrace{(n-4)(n+4)}_{\text{multiple of }n+4} + 16$

So the remainder is the same as the remainder when $16$ is divided by $n+4$
This will be $16$ if $n+4\gt16$ so if $n\gt12$.

Check values for $n=8$ to $n=12$ ($n$ up to $7$ shown above)
$n$ $n+4$ remainder when $16$ divided by $n+4$
8 12 4
9 13 3
10 14 2
11 15 1
12 16 0
So possible remainders are 0 - 7 and 16, 9 possibilites.



This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.

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