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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!

Filip Vlaisavljević
  • 438

2 Answers2

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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 : $(2c+1)^2=8\cdot\dfrac{c(c+1)}2+1\equiv1\pmod8$

lab bhattacharjee
  • 279,016
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We take those three type of inputs, because they cover all remainders on division by 4 ( odd covers 1 and 3 mod 4, even without being divisible by 4 is 2 mod 4, and divisibility by 4 is 0 mod 4). They also produce expressions we can easily find the remainder of on division by 8.