# $-a+b+c$ is divisible by 5, how about $a^2+b^2+c^2$?

Assume all these numbers are positive integers and the followings:

• $$-a+b+c$$ is divisible by 5,
• none of $$a$$, $$b$$ and $$c$$ are divisible by 5,
• $$a$$ is even and the other two are odd numbers.

How can we show that $$a^2+b^2+c^2$$ is not or is divisible by 5?

Expanding will result in $$-ab-ac+bc=5h$$ for some integer $$h$$. But then I cannot find other ways to dig more.

Another approach is to get the reminder of $$a$$, $$b$$ and $$c$$ and assume a similar problem, but this time $$0 and $$-a+b+c=0$$, which seems to be easier to deal with, but still I could not continue.

• Have you made any attempt, if so then please update it here. No one here is interested in doing your homework. Aug 11, 2021 at 8:46
• Thanks guys. This is not a homework, but rather a question of my own. I will update the question with some information. Aug 11, 2021 at 8:47
• "How can we show that $a^2+b^2+c^2$ is not or is divisible by 5?" Isn't that trivially true? Aug 11, 2021 at 9:01
• I mean is it true that this is not divisible by 5? and if not, under what conditions it is divisible by 5. Aug 11, 2021 at 9:02

The mod-5 residue of $$n^2$$ is either $$0$$ or $$±1$$. For $$a^2+b^2+c^2\equiv0\pmod 5$$ at least one of the three must be a multiple of $$5$$.

$$a=b=c=5 \to a^2+b^2+c^2=3\cdot 5^2$$ is divisible by 5

$$a=b=1, c=5 \to 1^2 + 1^2 + 5^2=27$$ is not divisible by 5

[UPD] if none of the number is divisible by 5: $$-a+b+c\equiv 0 \pmod 5 \to a\equiv b+c \pmod 5$$ Using $$\gcd(2, 5)=1$$ $$a^2+b^2+c^2\equiv 2bc+2b^2+2c^2\equiv 0 \pmod 5 \iff bc+b^2+c^2\equiv 0 \pmod 5$$

Three cases:

• $$b\equiv \pm1 \pmod5, c\equiv\pm1\pmod5$$ then $$bc+b^2+c^2\equiv \pm1+1+1\not\equiv 0 \pmod 5$$
• $$b\equiv \pm2\pmod5, c\equiv\pm2\pmod5$$ then $$bc+b^2+c^2\equiv \pm4+4+4\not\equiv 0 \pmod 5$$
• $$b\equiv \pm1\pmod5, c\equiv\pm2\pmod5$$ then $$bc+b^2+c^2\equiv \pm2+1+4\not\equiv 0 \pmod 5$$

The conclusion it is not divisible by 5

• Thanks. I should have been more specific. I will update the question. All these trivial cases are excluded. Aug 11, 2021 at 8:49
• Rather than break it up into cases this way, we can look when $b \equiv c \mod 5$ we have $3b^2\equiv 0 \mod 5$, which means $b \equiv 0 \mod 5$, a contradiction. So now the other case when $b \not \equiv c \mod 5$ we can rewrite $b^2+bc+c^2=\frac{b^3-c^3}{b-c}$ and ask when is $b^3\equiv c^3 \mod 5$? Since there is no element of order $3$, $b\equiv c \mod 5$ and again we have a contradiction. Aug 11, 2021 at 9:20