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Let $a,b,c \in N$ with the property that that 7 divides $a^3+b^3+c^3$, but it doesn't divide $a+b,b+c,c+a$.

a) Show that 7 divides $(a-b)^2+(b-c)^2+(c-a)^2$.

b) Find an example of numbers $a,b,c \in N$ with these properties.

my idea:

So for point $(b)$ I found the triples $(a,b,c)=(4,5,7)$ that have the wanted property

for point $(a)$ I thought of using $a^3+b^3+c^3= \frac{1}{2}*(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)+3abc$

The wanted property is easy to show if at leat one of the numbers is divisible by 7.

I don't know what to do if none of them is divisible by 7.

Jon
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IONELA BUCIU
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  • The implication is that in order for the conclusion to be true, one of them has to be divisible by 7. So you don't have to do anything if none of them is divisible by 7, but instead have to show that "at least one of the numbers is divisible by 7". – Calvin Lin Jan 23 '25 at 18:58
  • You can show that $a^3 \equiv 1 \text{mod} 7$ giving you $a + b + c \equiv 0 \text{mod} 7$, as $abc$ is not $\equiv$ to $0 \text{mod} 7$. An example is 1,2,4 for a, b and c respectively. – Kavya Sahai Jan 23 '25 at 19:01
  • @KavyaSahai Note that $1^3 + 2^3 + 4^3=73$ is not a multiple of 7. – Calvin Lin Jan 23 '25 at 19:03
  • Hint: Look at the values of $n^3 \pmod{7}$ to conclude that $abc \equiv 0 \pmod{7}$. The rest follows. – Calvin Lin Jan 23 '25 at 19:05
  • But $(-1)^2 + (-2)^2 + (3)^2$ is multiple of 7, given by the fact that $1 + 2 + 4 \text{ is } \equiv 0 \text{ mod } 7$. And, for this condition it is necessary that $abc$ is not a multiple of 7. – Kavya Sahai Jan 23 '25 at 19:06
  • I found the triples (a,b,c)=(4,5,7) that have the wanted property - actually, you only found a single example, namely $(4,5,7)$. – Dietrich Burde Jan 23 '25 at 19:15
  • @KavyaSahai I'm pointing out that your counterexample does not satisfy the conditions of the question, namely "the property that that 7 divides $a^3+b^3+c^3$", which applies to the subpart of what OP is interested in. (In particular, the conditions that $ 7 \not \mid a+b$ etc also need to apply. Side note: these actually aren't necessary.) $\quad$ I agree that $ 7 \mid (a-b)^2 + (b-c)^2 + (c-a)^2$ does not imply $ 7 \mid abc$. EG $(1, 1, 1)$ is a more obvious counterexample. – Calvin Lin Jan 23 '25 at 19:17
  • I don't understand the close votes. – Martin Brandenburg Jan 24 '25 at 03:45
  • By the same method described in the linked dupe we deduce that one of $,a,b,c,$ is $\equiv 0\pmod{!7},$ which enables you to finish as you described, using $(a).\ $ Please ask only one question per post. – Bill Dubuque Jan 24 '25 at 07:43

2 Answers2

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the cube of any integer is congruent to $0$ or $\pm1\mod7$, so wlog $a\equiv0\mod7$ and $b^3\equiv1 \mod7$ and $c^3\equiv6\mod7$

then $b\equiv1,2,4\mod7$ and $c\equiv3,5,6\mod7$

from the assumption that $b+c$ is not divisible by $7$, following combinations are not possible $(b\equiv1, c\equiv6), (b\equiv2, c\equiv5), (b\equiv4, c\equiv3)$

and if you expand the expression $(a-b)^2 + (b-c)^2 + (c-a)^2$ and ignore terms involving $a$ (because these are divisible by $7$) one only needs to show $$S=2\cdot(b^2+c^2-bc)$$ (or $b^2+c^2-bc$) is divisible by $7$:

case $b\equiv1, c\equiv3\implies S\equiv 14\equiv0\mod7$

case $b\equiv1, c\equiv5\implies S\equiv 42\equiv0\mod7$

case $b\equiv2, c\equiv3\implies S\equiv 14\equiv0\mod7$

case $b\equiv2, c\equiv6\implies S\equiv 56\equiv0\mod7$

case $b\equiv4, c\equiv5\implies S\equiv 42\equiv0\mod7$

case $b\equiv4, c\equiv6\implies S\equiv 56\equiv0\mod7$

R.Jean
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Let's work in the field $\mathbb{F}_7$, so that "divisible by $7$" becomes "is zero", which is easier. The only cubes in $\mathbb{F}_7$ are $0,1,-1$. Hence, you cannot write $0$ as a sum of three non-zero cubes (for example, $1 + 1 + (-1) \neq 0$). Thus, if $a^3+b^3+c^3=0$, then one these cubes and hence one of the $a,b,c$ must be zero. Let's say that $a$ is zero. By assumption, $b+c \neq 0$. Hence, $$\begin{align*} 0 &= a^3+b^3+c^3 \\ & = \tfrac{1}{2} (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)+3abc \\ & = \tfrac{1}{2} (b+c)((a-b)^2+(b-c)^2+(c-a)^2) \\ \implies 0 & = (a-b)^2+(b-c)^2+(c-a)^2. \end{align*}$$

Martin Brandenburg
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