Prove that when x, y, z are intiger number x^2+y^2+z^2 is divisible by 3

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Jimmy44

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xy+z, yz+x, xz+y are divisible by 3. Prove number x^2+y^2+z^2 is divisible by 3

1. We have intiger x, y and z and numbers \(\displaystyle xy+z\), \(\displaystyle yz +x\), \(\displaystyle xz +y\) are divisible by 3. Prove that \(\displaystyle x^2 + y^2 + z^2\) is also divisible by 3.

Please help me with this one.
 
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I'm not sure if my solution is good.
I've pointed out that these three sums are with two terms: one product and one number \(\displaystyle (x, y, z)\). And we can see that every singular term in every sum has to give us remainder of \(\displaystyle 1\) or \(\displaystyle 2\), or both terms in every sum have to give us remainder of \(\displaystyle 0\). Thus every number \(\displaystyle x, y\) or \(\displaystyle z\) give us remainder of \(\displaystyle 1, 2\) or \(\displaystyle 0\) and two from these remainders have to be the same (only exeption is \(\displaystyle 0, 0\) and \(\displaystyle 0\)). So we have can see now that when two remainders equal \(\displaystyle 2\), then the last one have to equal \(\displaystyle 1\) and otherwise when two remainders equal \(\displaystyle 1\), then the last remainder have to equal \(\displaystyle 2\). Define remainders of numbers \(\displaystyle x, y\) and \(\displaystyle z\) as \(\displaystyle a, b\) and \(\displaystyle c\). Then we can see that we have five possibilities:
1) \(\displaystyle (a, b, c) = (1, 1, 2)\)
2) \(\displaystyle (a, b, c) = (1, 2, 2)\)
3) \(\displaystyle (a, b, c) = (0, 0, 0)\)
Possiblity when \(\displaystyle (a, b, c)=(0, 0, 1 or 2)\) doesn't exist, because minimum in the one sum we will have situation when product is divisible by \(\displaystyle 3\) and one number \(\displaystyle x\) or \(\displaystyle y\) or \(\displaystyle z\) is that one with remainder of \(\displaystyle 1\) or \(\displaystyle 2\). Then that sum is not divisible by \(\displaystyle 3\).


We can see that square of number with remainder of \(\displaystyle 2\) give us remainder of \(\displaystyle 1\). Suppose we let \(\displaystyle x\) is that one with remainder of \(\displaystyle 2\). Then
\(\displaystyle x\equiv 2 (mod 3)\), so
\(\displaystyle x^2\equiv 4 (mod 3)\)
\(\displaystyle x^2\equiv 1 (mod 3)\)


So squares of numbers x, y, z give us remainders of 1 (\(\displaystyle 1^2=1\)), so
\(\displaystyle x^2 + y^2 + z^2\equiv 1+1+1 (mod 3) \)
\(\displaystyle x^2 + y^2 + z^2\equiv 0 (mod 3) \)
 
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\(\displaystyle x\ divisible\ by\ 3 \implies x = 3i,\ where\ i\ is\ an\ integer.\)

\(\displaystyle y\ divisible\ by\ 3 \implies y = 3j,\ where\ j\ is\ an\ integer.\)

\(\displaystyle z\ divisible\ by\ 3 \implies z = 3k,\ where\ k\ is\ an\ integer.\)

\(\displaystyle x^2 + y^2 + z^2 = (3i)^2 + (3j)^2 + (3k)^2 = 9i^2 + 9j^2 + 9z^2 = 3(3i^2 + 3j^2 + 3k^2).\)

If you divide \(\displaystyle 3(3i^2 + 3j^2 + 3k^2)\)by 3, do you get an integer? Why?
 
Yes, because if we divide \(\displaystyle 3(3i^2+3j^3+3k^2)\) by 3 we will get \(\displaystyle 3i^2+3j^3+3k^2\) and this is an integer if \(\displaystyle i, j\) and \(\displaystyle k\) are integer. But \(\displaystyle x, y\) and \(\displaystyle z\) may be not divisible by 3, so why did you suppose that 3 devide these numbers?
 
Mayby my title is mistakable, sorry. Full text of this exercise is on the top. ?
 
You can do this without modular algebra, but the basic idea is still the same as in your proof.

What is given is:

\(\displaystyle x,\ y,\ z,\ i,\ j,\ k \in \mathbb N^+,\ xy + z = 3i,\ yz + x = 3j,\ and\ xz + y = 3k.\)

\(\displaystyle Let\ x = 3a - p,\ where\ a \in \mathbb N^+\ and\ p = 0,\ 1,\ or\ 2.\)

\(\displaystyle Let\ y = 3b - q,\ where\ b \in \mathbb N^+\ and\ q = 0,\ 1,\ or\ 2.\)

\(\displaystyle Let\ z = 3c - r,\ where\ c \in \mathbb N^+\ and\ r = 0,\ 1,\ or\ 2.\)

\(\displaystyle qr - p \in \mathbb Z\ and\ -\ 2 \le qr - p \le 4.\)

\(\displaystyle yz + x = 3j \implies (3b - q)(3c - r) + 3a - p = 3j \implies 9bc - 3br - 3cq + qr + 3a - p= 3j \implies\)

\(\displaystyle qr - p = 3(j - a + br + cq) \implies \dfrac{qr - p}{3} \in \mathbb Z \implies qr - p = 0\ or\ 3.\)

\(\displaystyle qr - p = 0 \implies qr = p \implies four\ cases:\)

\(\displaystyle p = 2 = q\ and\ r = 1,\ p = 2 = r\ and\ q = 1, p = q = r = 1,\ and\ p = q = r = 0.\)

Without loss of generality, these can be reduced to two cases:

\(\displaystyle p = 2 = q\ and\ r = 1,\ and\ p = q = r.\)

\(\displaystyle qr - p = 3 \implies qr = p + 3 \implies one\ case:\ p = 1\ and\ q = r = 2.\)

So there are effectively 3 cases to look at.

\(\displaystyle Case\ I:\ p = q = r\)

\(\displaystyle x^2 + y^2 + z^2 = (3a + p)^2 + (3b + q)^2 + (3c + r)^2 \implies\)

\(\displaystyle x^2 + y^2 + z^2 = (3a + p)^2 + (3b + p)^2 + (3c + p)^2 \implies\)

\(\displaystyle x^2 + y^2 + z^2 = 9a^2 + 6ap + p^2 + 9b^2 + 6bp + p^2 + 9c^2 + 6cp + p^2 \implies\)

\(\displaystyle x^2 + b^2 + c^2 = 3(3(a^2 + b^2 + c^2) + 2p(a + b + c) + p^2)\)

\(\displaystyle \dfrac{x^2 + b^2 + c^2}{3} = 3(a^2 + b^2 + c^2) + 2p(a + b + c) + p^2 \in \mathbb N^+.\)

\(\displaystyle Case\ II:\ p = 1\ and\ q = r = 2.\)

\(\displaystyle x^2 + y^2 + z^2 = (3a + a)^2 + (3b + q)^2 + (3c + r)^2 \implies\)

\(\displaystyle x^2 + y^2 + z^2 = (3a + 1)^2 + (3b + 2)^2 + (3c + 2)^2 \implies\)

\(\displaystyle x^2 + y^2 + z^2 = 9a^2 + 6a + 1 + 9b^2 + 12b + 4 + 9c^2 + 12c + 4 \implies\)

\(\displaystyle x^2 + b^2 + c^2 = 3(3(a^2 + b^2 + c^2) + 6(a + 2b + 2c) + 3)\)

\(\displaystyle \dfrac{x^2 + b^2 + c^2}{3} = 3(a^2 + b^2 + c^2) + 6(a + 2b + 2c) + 3 \in \mathbb N^+.\)

\(\displaystyle Case\ III:\ p = 2 = q\ and\ r = 1.\)

\(\displaystyle x^2 + y^2 + z^2 = (3a + a)^2 + (3b + q)^2 + (3c + r)^2 \implies\)

\(\displaystyle x^2 + y^2 + z^2 = (3a + 2)^2 + (3b + 1)^2 + (3c + 1)^2 \implies\)

\(\displaystyle x^2 + y^2 + z^2 = 9a^2 + 12a + 4 + 9b^2 + 6b + 1 + 9c^2 + 6c + 1 \implies\)

\(\displaystyle x^2 + b^2 + c^2 = 3(3(a^2 + b^2 + c^2) + 6(2a + b + c) + 2)\)

\(\displaystyle \dfrac{x^2 + b^2 + c^2}{3} = 3(a^2 + b^2 + c^2) + 6(2a + b + c) + 2 \in \mathbb N^+.\)
 
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Jimmy44, I can't stress enough that you should try this proof because of its relative
brevity of lines and characters per line. So, I am recommending this one I referred
to in an earlier post instead of what you are doing:

xy + z, yz + x, xz + y are each divisible by 3. Prove number x^2 + y^2 + z^2 is divisible by 3.

Let xy + z = 3k

Let yz + x = 3m

Let xz + y = 3n


where x, y, z, k, m, n, belong to the complete set of the integers

Multiply each equation above by a corresponding needed variable:

\(\displaystyle xyz + z^2 = 3zk \ \ \ (1)\)

\(\displaystyle xyz + x^2 = 3xm \ \ \ (2)\)

\(\displaystyle xyz + y^2 = 3yn \ \ \ (3)\)


And now we can add (1) + (2) + (3):


\(\displaystyle 3xyz + x^2 + y^2 + z^2 = 3zk + 3xm + 3yn\)

\(\displaystyle x^2 + y^2 + z^2 = 3zk + 3xm + 3yn - 3xyz\)

\(\displaystyle x^2 + y^2 + z^2 = 3(zk + xm + yn - xyz)\)​


So, \(\displaystyle \ \ x^2 + y^2 + z^2 \ \ \) is divisible by 3.
 
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