195073

Q1) Every odd square leaves remainder 1 when divided by 8:
Odd squares do not have a factor of 2 - this means their square root must too be odd.
This means we can express an odd square as (2n-1)^2, which expands to 4n^2 - 4n + 1.
Now when we divide this by 8, we can separate it into two parts: (4n^2 - 4n)/8 + 1/8.
Clearly the first part can simplify to (n^2 - n)/2. From there we can factorise n out from the numerator to form (n(n-1))/2 - and as n is necessarily an integer, one of n and n-1 are even and the other is odd. Whenever an odd number is multiplied by an even one, the product is even - therefore (n(n-1))/2 is an integer. We are therefore left with only the 1/8 to consider - the remainder of 1.
Q2) Every even square leaves a remainder of 0 or 4 when divided by 8
Even squares have a factor of 2 - therefore their square roots are even.
This means even squares can be expressed as (2n)^2, which expands to 4n^2.
Now if n^2 is odd, 4n^2 does not gain another factor of 2 to be a multiple of 8, but is still a multiple of 4; meaning that dividing it by 8 gives a remainder of 4.
On the other hand, if n^2 if even, 4n^2 gains at least another factor of 2x2 = 4, and becomes a multiple of 4x4 = 16, and therefore also a multiple of 8 - yielding a remainder of 0.
Q3) Any 8n + 7, where n is a positive integer, cannot be the sum of three squares
8n is even and 7 is odd - the sum of an even and odd number is another odd number.
Now if we assume the first square is even, the next two have to sum to an odd number so that the total sum is odd. This can only happen with one more even square and an odd square.
If we assume the first square is odd, the next two have to sum to an even number so that the total sum is odd. This can only happen with either two more even squares (giving us the same case as if the first is even) or with two more odd squares (a unique case).
Therefore we have to consider two cases; where there are two even squares and one odd square, and where there are three odd squares.
Let's consider the case with three odd squares first. Odd squares can be expressed as (2n-1)^2, which expands to 4n^2-4n+1. This means that with three odd squares, we can write the sum as 4x^2 - 4x + 1 + 4y^2 - 4y + 1 + 4z^2 - 4z + 1, which can be simplified and factorised to give 4(x^2 - x + y^2 - y + z^2 - z) + 3, and factorised further to give 4(x(x-1) + y(y-1) + z(z-1)) + 3. Now we can write this expression as an equation being equal to 8n+7:
4(x(x-1) + y(y-1) + z(z-1)) + 3 = 8n + 7 [subtract 3 from both sides]
4(x(x-1) + y(y-1) + z(z-1)) = 8n + 4 [divide by 4]
x(x-1) + y(y-1) + z(z-1) = 2n + 1
Now as 2n + 1 is odd (even+odd), x(x-1) + y(y-1) + z(z-1) must be odd - however as x, y and z are all integers, one of x and x-1 is odd and the other is even, meaning that the product is even. This logic also applies for y(y-1) and z(z-1), meaning that it is a sum of three even numbers, yielding another even number. This is a contradiction, so therefore three odd squares cannot sum to 8n+7.
Now let's consider the case with one odd square and two even squares. An even square can be expressed as (2n)^2, which expands to 4n^2, and odd squares can expand to 4n^2 - 4n + 1 as shown earlier. Therefore this sum can be expressed as 4x^2 - 4x + 1 + 4y^2 + 4z^2, which can be factorised to give 4(x^2 - x + y^2 + z^2) + 1. This expression can now be written in an equation as being equal to 8n+7:
4(x^2 - x + y^2 + z^2) + 1 = 8n + 7 [subtract 1 from both sides]
4(x^2 - x + y^2 + z^2) = 8n + 6 [divide by 4]
x^2 - x + y^2 + z^2 = 2n + 6/4
It can clearly be seen that 2n + 6/4 is not an integer if n is an integer. However, as, x, y and z are all also integers, x^2 - x + y^2 + z^2 must also be an integer - this is a contradiction, therefore two even squares and an odd square cannot sum to give 8n + 7.
As neither of the cases can be equal to 8n+7, there is therefore no set of three squares which sums to give any 8n + 7.