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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!

IONELA BUCIU
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  • The term "pure algebra answer" doesn't have a definition. – WhatsUp Nov 05 '23 at 13:29
  • Hello :) There are only two examples with a value less than 28: $(2, 3,5) $ and $(3, 5,7) $. Wlog let $a<b<c$.If $a>3$, then $c-a\geq 6$, because $a, b, c$ are $\pm 1$ mod 6. So, there are at most 2 primes in ${a,\cdots a+5}$ for $a>3$. If $c-a=6$, then ${b-a, c-a, c-b}={2, 4,6}$ and then $(b-a) ^2+(c-a) ^2+(c-b) ^2=56$. – Jochen Nov 05 '23 at 13:39
  • I haven't given this any thought to see whether it could help, but I notice that $E$ is one of the two factors of the factorization of $a^{3} + b^{3} + c^{3} - 3abc$ into a product of non-constant polynomials with integer coefficients. – Dave L. Renfro Nov 05 '23 at 13:49
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    @Jochen And there are $(2,3,7)$ with $E=21$ and $(2,5,7)$ with $E=19.$ – Reinhard Meier Nov 05 '23 at 13:58
  • Hello :) @ReinhardMeier You are right, sorry. It is my mistake. – Jochen Nov 05 '23 at 14:00
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    Just to amplify Jochen's comment: If none of the primes is $2$ or $3$, then the closest primes can be clustered is $p,p+2,p+6$ or $p,p+4,p+6$, each of which fails the required conditions. Primes that include larger gaps fail by a larger margin. Therefore, collections of primes that satisfy the conditions must include $2$ or $3$ or both. Those collections have been identified in other comments. – Keith Backman Nov 05 '23 at 17:02
  • I would have thought there was enough information in the comments for you to put together a complete answer on your own, IONELA. Have you tried? – Gerry Myerson Nov 15 '23 at 11:31
  • @GerryMyerson Yes, i have tried, but i dont understand why if none of the primes is 2 or 3, then the closest primes can be clustered is p,p+2,p+6 or p,p+4,p+6 – IONELA BUCIU Nov 15 '23 at 11:38
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    If the difference between largest and smallest primes is five, then one of those two primes must be even, but the only even prime is two. So, if no prime is two, the largest and smallest primes can't differ by five. If the difference between largest and smallest is four, then, since all three primes must be odd, they must be $p,p+2,p+4$ for some prime $p$. But given any three numbers $n,n+2,n+4$ at least one of those numbers must be divisible by three. If that number isn't the prime three, but it's divisible by three, then it must be composite. So if three isn't used, (continued) – Gerry Myerson Nov 15 '23 at 11:53
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    (continued) then biggest minus smallest can't be four. Similar reasoning rules out biggest minus smallest equals three (excepting $2,3,5$) and biggest minus smallest equals two. – Gerry Myerson Nov 15 '23 at 11:55

3 Answers3

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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.

Piquito
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  • It's a standard conjecture, that there are infinitely many primes $p$ such that $p+2$ and $p+6$ are both prime; also, infinitely many primes $p$ such that $p+4$ and $p+6$ are both prime. – Gerry Myerson Nov 17 '23 at 03:53
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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}

Claudio
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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 $$\color{red}{(a,b,c,E)=(2,3,5,7),(2,3,7,21),(2,5,7,19),(3,5,7,12)}$$

mathlove
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