What numbers smaller than $28$ can be values of the expression $E=a^2+b^2+c^2-ab-bc-ca$?

Question

question

What numbers smaller than $28$ can be values of the expression $E=a^2+b^2+c^2-ab-bc-ca$, where $a<b<c$ are prime numbers?

my idea

We have to solve the following inequality

$a^2+b^2+c^2-ab-bc-ca=28 |*2$

$2a^2+2b^2+2c^2+2ab+2ac+2bc<56$

$(a-b)^2+(b-c)^2+(a-c)^2<56$

because $a<b<c$, $(a-c)^2$ is the greatest number in the inequality.

Also, because the numbers are perfect squares, we can say that $0<(a-b)^2+(b-c)^2+(a-c)^2<56$

I know that from now I can simply start doing tries, but I am looking for pure algebra.

Hope one of you can help me! Thank you!

Answer 1

You have correctly said that $(c-a)^2$ is the greatest number in the LHS of $(b-a)^2+(c-b)^2+(c-a)^2<56$.

Now since $$(c-a)^2\lt (b-a)^2+(c-b)^2+(c-a)^2\lt 56$$ we have to have $(c-a)^2\lt 56$ which implies $c-a\lt 8$.

Let us separate it into two cases :

If $a=2$, then it follows from $c-a\lt 8$ that $c\le 7$ where $c$ is a prime number. So, $(a,b,c)=(2,3,5),(2,3,7),(2,5,7)$.

If $a\ge 3$, then $a,b,c$ are odd. So, $b-a,c-b,c-a$ are even. Since $2\le b-a\lt c-a\le 6$ and $2\le c-b\lt c-a\le 6$, we have three cases to consider.

• Case 1 : $(b-a,c-b,c-a)=(2,2,4)$
  Suppose that the remainder is $1$ when $a$ is divided by $3$. Then, $b=a+2$ is divisible by $3$. Since $3$ is the only prime number divisible by $3$, we have to have $b=3$, and then $a=1$ which is not a prime number. So, the remainder is not $1$. Suppose that the remainder is $2$ when $a$ is divided by $3$. Then, $c=a+4$ is divisible by $3$. So, $c=3$, and then $b=1$ which is not a prime number. So, the remainder is not $2$. Therefore, the remainder has to be $0$ when $a$ is divided by $3$, which means that $a$ has to be $3$. So, $(a,b,c)=(3,5,7)$.

• Case 2 : $(b-a,c-b,c-a)=(2,4,6)$
  We have $(b-a)^2+(c-b)^2+(c-a)^2=56$ which contradicts that $(b-a)^2+(c-b)^2+(c-a)^2\lt 56$.

• Case 3 : $(b-a,c-b,c-a)=(4,2,6)$
  We have $(b-a)^2+(c-b)^2+(c-a)^2=56$ which contradicts that $(b-a)^2+(c-b)^2+(c-a)^2\lt 56$.

Therefore, the only solutions are $$\color{red}{(a,b,c,E)=(2,3,5,7),(2,3,7,21),(2,5,7,19),(3,5,7,12)}$$

Answer 2

From $E=a^2+b^2+c^2-ab-bc-ca < 28 $ it follows that $(b-a)^2+(c-b)^2+(c-a)^2<56$, where $a<b<c$ are prime numbers. As mentioned by you, $(c-a)^2$ is the greatest number in the inequality and therefore $(c-a)^2 < 56$ or $c-a < 8$. Given that min $a$ = 2, max $c$ = 7 (under the assumption that $a,b,c$ are prime numbers) and $b \in \{3,5\}$.

Hence $a \in \{2,3\} \land b \in \{3,5\} \land c \in \{5,7\}$ and the possible solutions are

\begin{align} \{a, b, c, E\} = \{2, 3, 5, 7\} \lor \{2,3,7,21\} \lor \{2,5, 7, 19\} \lor \{3, 5, 7, 12\} \end{align}

Answer 3

COMMENT.-$(1)$ Contrary to the appearance, it seems to me there are infinitely many solutions maybe. For example, we have, being $f(a,b,c)$ the given expression, $$f(47,53,49)=f(47,43,49)=f(47,41,45)=f(47,51,45)=28$$ (note that these are not examples because $49,45$ and $51$ are not prime but it allows us to see integer solutions for numbers greater than $28$)

Besides, for all fixed value of one variable , say $a$ we have the equation of an ellipse which could have integer coordinates (to follow).

$(2)$ Two examples with large primes
$$f(613,617,619)=28\\f(641,643,647)=28$$

There are many integers such that $f(a,b,c)=28$, two of them being prime. The curious thing is that very frequently there are integer solutions and it seems clear that there are an infinite number of integer solutions. My feeling is that there is also an infinity with $a,b,c$ primes.