My mistake, I multiplied both sides of the equation by 0.
The two cases are N>2 and N<=2, for N being real, X should be positive. This is a problem with polynomials over the real field; sometimes there are no roots...
Very nice code Anza Power, thank you for writing it. It would indeed cover all possibilities, and checks that they sum to 34. It is the correct figure provided the code compiled correctly, and that the last 'for loop' ranges from C+1 to 16 only (but the result posted were ran with the correct value). I would like to make an apology.
I had in mind using the magic squares of order 4 to answer this question. It was not well thought out in that even though I recognized the fact there being 220 fundamentally different magic squares of order 4, I did not confirm that they may contain the same sums between more than one of them. However, this question can be properly phrased if 34 were replaced with 15, 16 replaced with 9, and 4 replaced with 3. This is because there is only one magic square of order 3, barring rotations and reflections. My apologies for so hastily posting this question without confirming that a fundamental square only contains their own unique sums, which it does not.
For 1<=A<B<C<D<=9=3^2, such that A+B+C+D=15, there are 8 ways.
Thank you all for their reply or thinking on this topic, and working so diligently to solve it.
What will bring one merit beyond the number of sands in the Ganges river, when each grain of sand in the Ganges river contained a Ganges river with its banks of sands? That which will bring you upon understanding even 4 lines will bring your supreme clarity and awakenedness?
I think the best we can do is prove that 1 + 1 = A, is unique. That is to say, only the element A comes from 1 + 1.
Let, 1 + 1 = A and
suppose that 1 + 1 = B, B not equal to A. So now there is another element that comes from 1 + 1 and is not A.
Then (1 + 1)+(1 + 1) = A + A = (1 + 1)+(1 + 1) = A + B
So A + A = A + B
(A + A) - A = (A + B) - A [i am not sure about this step]
A + (A - A) = (A + B) - A
A + 0 = (B + A) - A
A = B + (A - A)
A = B + 0
A = B
and we arrive at a contradiction.
From now on, we will write 2 instead of A, so that 1 + 1 = 2. So whenever we see 1 + 1, we can say it equals to 2.