A number theory problem by Áron Bán-Szabó

There will be two numbers from $x$ , $y$ and $z$ which are both even or both odd ( Pigeonhole Principle ). Let $x$ and $y$ be these numbers. Then $(x+y)$ is an even number, so $(x+y)*(x+z)*(y+z)$ has to be even, but $2017^n$ will never be an even number. So it is not possible.

• Since $2017$ is a prime,

• $x + y = 2017^i$

• $y + z = 2017^j$

• $z + x = 2017^k$

• Solving above equations,
• $x = \frac{2017^i + 2017^k - 2017^j}{2}$
• Since $2017^i + 2017^k - 2017^j$ is not even, x is not an integer.

2017 is a prime number. Therefore all factors of $2017^n$ are powers of 2017 and therefore odd. Now $2x = (x+y) + (x+z) - (y+z) = 2017^a + 2017^b - 2017^c$ is odd, so $x$ cannot be an integer.

$( x + y ) + ( y + z ) + ( z + x ) = 2( x + y + z ),$ and this implies that one or all of $( x + y ), ( y + z ),$ and $( z + x )$ must be even. $( x + y )( y + z )( z + x )$ is even, but $2017^n$ is odd, which gives the answer that this is not possible.

Assume for the sake of contradiction, there exists a minimal solution $(x_0,y_0,z_0)$ of positive integers that satisfy this equation. Since $2017$ is prime, it must divide every factor on the left hand side, that is $x+y\equiv x+z\equiv y+z\equiv 0$ modulo $2017$ . Moreover, this equation implies that $x-y, y-z,$ and $z-x$ are multiplies of $2017$ . The last two equations imply that $2017$ divides each of $x,y,z$ , so a smaller solution $(x',y',z')=\frac{1}{2017}(x,y,z)$ satisfies the original problem's equation (since $x,y,z$ are positive), thus contradicting minimality.

By infinite descent, there is no solution to this equation.

As 2017 is prime no. Ergo, there can't be any solution to it.

2017 is prime. Therefore, if three numbers multiply together to give some power of 2017, n is greater than or equal to 3, so that rules out n = 1 and n = 2 as possibilities. If you try n = 3, there are a few systems of linear equations to test out:

x + y = 2017 and y + z = 2017 and x + z = 2017

OR

x + y = 2017 ^ 2 and y + z = 2017 and y + z = 1

OR

x + y = 2017^3 and x + z = 1 and x + z = 1

And all of them yield non-integer answers for x, y, and z. Thus the problem is not possible.

The sum of two integers is an integer. Therefore x + y, x + z, y + z will be three integers.

Since 2017 < 2700 (30 * 30 * 30) we really need only test numbers below 30 but prime factorization of 2017 below 30 is:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

The numbers which make up the three numbers will either be prime or made of primes below 30.

I'm not sure how to express this logic or prove it, but that was how I solved it, without fully proving it is prime. There cannot be three numbers if it is prime.