There are only 8 values for N^2 mod28: 0, 1 , 4, 8, 9, 16, 21, 25.

To make the amount a^2+b^2+c^2 equal to 28*k (k a non-negative integer) we will choose a triplet (a ,b, c) fulfilling a+b+c =0 mod 28.

It is obtainable by numbers whose mod28 representation is (a,b,c) = (0, 0, 0) or (4, 8, 16).

5 numbers correspond to 0: 14, 28, 42, 56, 70 (each of those squared is divisible by 28) (i)

10 numbers correspond to 4: 2,12,16,26,30,40,44,54,58,68 (each of those squared is 28k+4) (ii)

10 numbers correspond to 8: 6,8,20,22,34,36,48,50,62,64 (each of those squared is 28k+8) (iii)

10 numbers correspond to 16: 4,10,18,24,32,38,46,52,60,66 (each of those squared is 28k+16) (iv)

(0, 0, 0) combination yields choice of 3 numbers from the (i) set i.e. * 5+10+20=***35 choices**

(4, 8, 16) combination - choice of 3 numbers, one each of the (ii), (iii), (iv) sets

i.e. 10^3=**1000 choices**

Answer: there are *1035* possible solutions.

Hope I did not err.

*Apparently I did: overcounted - I considered triplets 111,112,113,...555 without noticing that 112,121,211 and similar count as one , not three. I got 5^3=125.- ***wrong thinking**

*The correct number is 35 (5 by using the same number e.g. 14,14,14 ;*

20 -one repeated twice and the third one of the remaining four e.g. 14,14,28; * *

10-three distinct numbers e.g. 14,42,56).

*So the correct answer is 1035.*

**Thanks due to Dej Mar for pointing out my error.**

BTW, the creation of list like **4,10**,18,24,32,38,46,52,60,66 is extremely easy :

1. First you evaluate n^2 MOD 28 for n=0

to 7, since (14-n)^2=n^2 taken MOD 28 .

2. Find the n (within the range 0-7)that corresponds to **16**(for OUR example), get 4.

3, Clearly 14-4=10 is the next in the list, so we have a couple (4,10) to begin with .

4. Add to this couple (14,14), get (**4,10**,18,24,)

5. The extension of this set is made by adding **14** to each of the two last members, ...and again not go over 70.

- piece of cake.

*Edited on ***December 22, 2014, 12:51 am**