I've been practising Math and I'm now stuck at the following exercise. I have the solution written down:
Exercise 1.
Sum of three square of integers when divided by 8 can't leave a remainder of 7. Prove it.
Lets write these three numbers as following:
Odd number : $(2k+1)^2 = 4k^2+4k+1 = 4k(k+1)+1 = 8k+1$
Even number:$(4k^2)=16k^2$
Even but not divisible by 4:$ (4k+2)^2 = 16k^2+16k+4$
So we get a set of the numbers {0,1,4).
0 = 0+0+0
1 = 0+0+1
2 = 0+1+1
3 = 1+1+1
4 = 0+0+4
5 = 0+1+4
6 = 1+1+4
7 -> can't be written by the set of the numbers. So, its proved.
My questions is, whats unclear, why do we take these three numbers as one odd, one even and one even but not divisible by 4?
Thanks in advance!
Best Answer
Think of $\pmod2,$ there can be $2$ possible in-congruent residues
For $\pmod4,$ its $4$
So, any integer $b\in[4a-1,4a,4a+1,4a+2]$ where $a$ is any integer
Now
$(4a+2)^2\equiv?\pmod8,$
$(4a)^2\equiv?\pmod8,$
$(4a\pm1)^2=16a^2\pm8a+1\equiv1\pmod8$
$4a\pm1$ have been correctly grouped as odd numbers
Another way for odd numbers : $(2c+1)^2=8\cdot\dfrac{c(c+1)}2+1\equiv1\pmod8$