Prime Triangle

Let $x,y,z$ denote three distinct positive integers and $p_1, p_2,p_3$ be the prime numbers, such that $\begin{array}{rl} x + y &= p_1\\ y + z &= p_2\\ x + z &= p_3 \end{array}$ Combining all these equations altogether, we see that $2\left(x + y + z\right) = p_1 + p_2 + p_3$ As one of the solutions from the previous problem notes, if all prime numbers are odd, then there is no way to achieve the sum.

Otherwise, at least one prime is even (which is clearly the number $2$ ) to make the above equation even, which conflicts with the assumption that all integers are distinct and positive. Therefore, it is impossible to have $p_1 + p_2 + p_3$ be even.

Thus, there is ${\color{#D61F06}\boxed{\text{no way}}}$ to construct three prime sums.

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Extension Problem: I have posted the generalization problem of mine related to these two posted earlier. There is in fact the pattern for determining whether the construction is possible or not.

Have fun trying out my generalization problem :)

First note that if the three vertex numbers are distinct, all three pair-wise sums (sums along the edges) are distinct. Also remember that there is only one even prime (the number 2). Bearing these two facts in mind we will show that a solution is impossible.

We will consider three cases which cover all possibilities.

First suppose that all the vertices are odd. Then all the pair-wise sums are odd+odd=even. Then a solution would give us three distinct even primes which is impossible.

Secondly suppose there are at least two even vertices. Then at least one of the pair-sums is even+even= even. Moreover this even number is greater than 2, and so cannot be a prime. So this case is also impossible

The remaining case is to have one even and two odd vertices. The odd vertices add to give an even number, and this will be a prime only if both vertices are = 1. But this breaks the rule that all the vertices are distinct. So this last case is also impossible.

Done.

First off, for the three circles in the diagram, let's denote the values A, B, and C, and for the primes, let's denote them as $p_1$ , $p_2$ , and $p_3$ .

Secondly, the problem itself is really asking if we can satisfy this system of equations :

$\begin{cases} A + B = p_1\\ B + C = p_2\\ C + A = p_3 \end{cases}$

The very first step to figuring that out is first considering that all prime numbers except 2 are odd numbers. Now, consider that two odd numbers added together always equals an even number and that one odd plus one even is always odd:

1. $\color{#D61F06}Odd$ + $\color{#D61F06}Odd$ = $\color{#3D99F6}Even$
2. $\color{#D61F06}Odd$ + $\color{#3D99F6}Even$ = $\color{#D61F06}Odd$

Considering that all sums like statement 1 will result in an even number greater than 2 given the conditions of the problem, it is safe to say that so far, it is impossible to have the system above satisfied.

It is much trickier when we consider sums like statement 2, though once again, let's consider our system of equations:

$\begin{cases} A + B = p_1\\ B + C = p_2\\ C + A = p_3 \end{cases}$

Since we are looking for an odd number for values $p_1$ , $p_2$ , and $p_3$ , Let's say A is an even number:

$\begin{cases} Even + B = p_1\\ B + C = p_2\\ C + Even = p_3 \end{cases}$

At this point, the only possible way to make \(p_1 a prime, and p_2 a prime is by making B and C odds, assuming both sums already make a prime number. Now comes the second case in the system of equations, if we already said that B and C are both odds, that raises a red flag that p_2 will end up being even, and greater than 2. We will come to this same predicament regardless of what value A, B, or C, we choose to be even.

\(\begin{cases} A + Even = p_1\\ Even + C = p_2\\ C + A = p_3 \end{cases}\)

$\begin{cases} A + B = p_1 \\ B + Even = p_2 \\ Even + A = p_3 \end{cases}$

Therefore, it is $\boxed{impossible}$ to fill in the diagram with the set conditions.

It is clear that it is not possible, to add two distinct positive integers to get 2. So if it is possible, then the sum of any two numbers is an odd prime number, because 2 is the only even prime number. But that's not possible, because there will be two numbers, which are both even or both odd, so the sum of them is even.

if you consider 0 as positive, you could have 0,2,3.

In any set of 3 positive numbers, either two of them will be even or two of them will be odd (Pigeonhole Principle). In both cases, you'll have one pair of numbers that sum to an even number. In order for the even sum to be prime, the sum of this pair would have to be 2 (since the only even prime is 2). But it is impossible to add two, different, positive numbers to get 2, so the task before us is impossible.

Other than "2" all prime numbers are Odd numbers. (odd + even =odd, odd + odd =even & even + even=even)

we are not able to have this kind of combination for distinct three positive integers.(odd + even =odd)

So It is not possible the pair of adjacent circles adds up to a prime number.

The only prime even number is 2, any other even number would be divisible by two. This means that the only way this could be solved is if you could use 1 twice, but that's not possible. So, you'll either use 2 even numbers or 2 odd numbers (other than 1) which will result in an even number

Fact 1: The sum of two distinct odd numbers is even (and not prime). Fact 2: The sum of two even numbers is even (also not prime).

We need to show that, in every possible case, at least one of the three pairwise sums will be composite. Let's examine all possible cases. There are four cases: I) All odd, II) all even, III) 2 even & 1 odd, and IV) 1 even & 2 odd.

Case I fails by Fact 1. Case II fails by Fact 2. Case III fails because the pairwise sum with both even numbers is necessarily even (i.e. Fact 2 again). Case IV fails because the pairwise sum with both odd numbers is necessarily even (i.e. Fact 1 again).

None of the four cases have it so that all three pairwise sums are prime. Therefore, it is impossible to fill in the circles such that each pair of adjacent circles adds up to a prime number.

Ok, I'm not a mathematician like some of you guys so I had to brute force logic lol. Here's how I solved it:

OK I confess, I know some math. Enough to be nearly dangerous, like an old lady with a skillet. So here are my Axioms:

1.) All Prime Numbers are odd (Except that little devil # 2)

2.) (<- There he is, that piece of -) Odd numbers can be expressed as (2^n)-1 for n >= 0 or (2^n)+1 for n > 0 where n is a real integer. Why is the second one for n > 0? Because it would define 2 as odd (which I think it is! It fits the description of being only divided by itself and 1 but semantics and what-ever don't need angry nerds kicking down my door today)

3.) Here's a cool one. Historian and Amateur Mathematician Christian Goldbach discovered (yet unproven, though I feel like it can easily be shown) that all even numbers greater that 2 can be shown as the sum of two primes. This conjecture is important to my solution.

Ok so because of #1 we know that only odd plus an even (2+3 = 5. Ok so maybe 2 IS even. Great...) will get us a prime. Why? well if we add two primes expressed as (2^n) + 1 (see #2 if irked about 2 being odd) we see that ((2^n)+1) + ((2^n)+1) => (2^n)+(2^n) +1+1 = (2^n)+(2^n) + 2 which is obviously even. I mean look at all those twos (That piece of-)! Also we know we can take the log base 2 of this sum and get a nice integer proving it to have a divisor of 2. Now if we add an even and an odd: (2^n) + (2^n)+1 => (2^(n+1)) +1 => (2^m) + 1 which is odd since it fits axiom #2. (But Shawn that doesn't mean it's a Prime) YES I KNOW THAT POINT DEXTER! It just states the * possibility * that is is prime and it can only be a prime when the sum is odd (see #1).

So we know two odds equal a even. So the only way we can construct this triangle of love... Ehem, of numbers is to have an odd number, some weird quasi even number like 2 (You piece of -! THAT'S RIGHT PEOPLE! Comedy comes in Threes!) and an even number to prevent two odds from adding up. BUT WAIT! 2 is a * real * work of art, and we know that in this triangle it will add up with that other even number which would be (2+ (2^n)) which we know will be even (LOOK AT ALL THOSE TWOS!) and thus not odd and thus not prime. So we logically can't do this.

Since there are 3 different numbers we know that they can be: odd, odd, even or even, even, odd. Any other combination will lead to there being more than one even sum which means two different even primes which is impossible. We know that one of the sums must be even which means one of sums must be the only even prime 2.

Therefore the even, even, odd combination must contain 2,0 to make the 2 and another prime to complete the trio. The prime is infact the lesser of any twin prime pair. However, of course 0 is not a positive integer. Hence this is not valid.

The other combination odd, odd, even, the odd, odd must add up to 2. But there does not exist two odd integers which are positive and less than 2. Hence there is no such combination.

This means the statement is false for any 3 distinct positive integers.

Don't over think it! All prime numbers are odd. Try to add any odd/even mix and get all odd numbers. Can't be done.

Let the three numbers be x, y and z. The three primes be p1, p2 and p3. p1, p2 and p3 cannot be even. The only even prime is 2. Because the numbers are needed to be distinct and positive. So p1, p2 and p3 is odd. x + y = p1. Since p1 is odd, then x is odd and y is even, or x is even and y is odd. y + z = p2. y is even and z is odd, or y is odd z is even. This will lead us to a resultant of 2 odd numbers out of x, y and z. Two odd numbers will add up to give an even number. As we know that, all even numbers except 2 are composite, thus, the above question will never have a solution.

Only one of the numbers in the corners can be even in order to have sums attached to it are odd and prime. It means that two other numbers are odd therefore their summation is even and is not prime. So the answer is NO

Two prime numbers must add up to an even number, which cannot be prime, since it's divisible by 2.

It is not possible if the numbers are distinct.

At least one pair of numbers will be both even or both odd.

Either way the sum of those two will be even, not prime.

The only way to get 3 prime sums is with 1,1,2 summing to 2,3,3 but that violates the requirement for distinct numbers.

Consider the parity of the three numbers at the vertices of the triangle:

Since the sum of two even integers is even, and the sum of two odd integers is also even, we can see that no matter the choices for parity of the three numbers, at least one sum must be even.

However, the only even prime number is 2 -- and since we cannot place the number 1 at two separate vertices (the numbers at the vertices must be distinct), there can be no solution.

a much easier approach. by hit and trial we deduce that 2 cannot be any of that positive integer and so two cannot be an answer .now as two cannot be the required prime two of those circles m ust contain odd positive integers but odd + odd =even and 2 as the only prime is not an answer ..results in a contradiction