marsupialsoup
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marsupialsoup added a post in a topic
jagdmc, you can use type sum n^2 in WolframAlpha, that it should give you the partial sum formula you can incorporate into your solution.

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marsupialsoup added a topic in New Logic/Math Puzzles

marsupialsoup added a post in a topic
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...

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marsupialsoup added a post in a topic

marsupialsoup added a post in a topic
Here is to solve for the dividend.
It gives the same answer as the other posters, I just wanted to make sure.

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marsupialsoup added a post in a topic
Only one solution for the divisor is possible, here is why,

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marsupialsoup added a post in a topic

marsupialsoup added a post in a topic

marsupialsoup added a post in a topic
We will use some tools from the internet to do the first part of this puzzle.

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marsupialsoup added a post in a topic
For that unique magic square of order 3, there are 8 ways to sum to 15, 8 = 3 rows + 3 columns + 2 diagonals.

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*That which will bring you upon understanding even 4 lines of it supreme clarity and awakenedness?
LJayden and plainglazed are both correct.

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marsupialsoup added a post in a topic
Mistake, please consider that for integers, 1<=A<B<C<=9, A+B+C=15 has 8 ways.

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marsupialsoup added a post in a topic
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.

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marsupialsoup added a topic in New Logic/Math Puzzles

marsupialsoup added a post in a topic
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.
Proof:
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.

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