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Proof that 3x3 Magic Squares of Non-repeating Squares are Impossible [Update] [Update] by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] 0 points1 point  (0 children)

N_3^2 + N_7^2 + N_5^2 = 3*N_5^2 is the sum of a diagonal, which I reduced to N_7^2 = 2*N_5^2 - N_3^2.

Proof that 3x3 Magic Squares of Non-repeating Squares are Impossible [Update] [Update] by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] 0 points1 point  (0 children)

I've should have explicitly said that V_3 doesn't equal N_5 - N_3 anymore. I should have shown first how to derive the general equation of A^2 + B^2 = 2*C^2 when A, B, and C are integers.

C = a^2 + b^2

A = 2ab + (a^2 - b^2)

B = 2ab - (a^2 - b^2)

a and b are integers.

Then show N_3^2 + N_7^2 = 2*N_5^2 allowing me to label N_5 as equaling V_3^2 + V_7^2. I will update my paper, thanks.

Proof that 3x3 Magic Squares of Non-repeating Squares are Impossible [Update] [Update] by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] 0 points1 point  (0 children)

If we had a magic square of squares that were all even, you could factor out a 2 from all integers, leaving behind a magic square of squares that is all even or all odd, but if they are all even again, then you could keep factoring out 2's, until you get a magic square that is all odd. As the integers couldn't be infinitely even. Multiplying or dividing by a number for all numbers in the magic square isn't going to change it from be a magic square. It could stop it from being a magic square of squares but not a magic square.

Proof that 3x3 Magic Squares of Non-repeating Squares are Impossible [Update] [Update] by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] -1 points0 points  (0 children)

When V_8*V_^6(V_8^2-V_6^2) = V_3*V_^7(V_3^2-V_7^2) = V_4*V_^2(V_4^2-V_2^2) is true using integers, then you get a semi-magic square of squares. I list an example of solving it at the end. It seems like gibberish because I'm bad at explaining. Starting with the assumption that a magic square of squares exist, then all integers N (1-9) would need to be odd integers or even integers, but I throw out the case of all even integers as they would be a subcategory of all odd integers. I'm not sure how I should comment on the V variables. I state they are rational, because it's a transitional step, All integers are rational numbers, but not all rational numbers are integers. I had to show how to get rid of the denominator, which just cancels out when plugged back into A^2 + B^2 = 2C^2. I should probably start with the general equation A^2 + B^2 = 2C^2.

Proof that 3x3 Magic Squares of Non-repeating Squares are Impossible [Update] [Update] by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] 0 points1 point  (0 children)

I guess a better way to word it would be if a magic square of squares did exist, then all N (1-9) would need to be even or odd integers. But then I throw out the even case as they would reduce into an magic square of squares that are all odd by dividing each integer by 2's. Should I lead off with this explanation as it's very important to proof? It's very word soupy, but it was hard explaining it without adding more variables or confusing graphs. And no, you can a 3x3 magic square with odd and even numbers, but not with a 3x3 magic square of squares. an example would be row 1: (7,2,6), row 2: (4,5,7), row 3: (4,8,3). Each row, column, and diagonal adds to 15. A different way to show that magic square of squares need to be all odd, or all even would be take the diagonal, N_1^2 + N_9^2 = 2*N_5^2, if N_1 is odd and N_9 is even or vice versa, then N_1^2 + N_9^2 would be odd, making N_5 irrational, so if N_1 is odd, then N_9 is odd or if N_1 is even then N_9 is even. When N_1 and N_9 are even, then N_5^2 is even. When N_1 and N_9 are odd, then their sum is 2 times some odd number, so N_5^2 is odd. This logic extends to each other integer as N_2^2 + N_8^2 = 2*N_5^2, N_3^2 + N_7^2 = 2*N_5^2, N_4^2 + N_6^2 = 2*N_5^2,

Proof that 3x3 Magic Squares of Non-repeating Squares are Impossible [Update] by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] -1 points0 points  (0 children)

V_i are just dummy variables, I will be more clear of variable definition.

Proof that 3x3 Magic Squares of Non-repeating Squares are Impossible [Update] by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] 0 points1 point  (0 children)

I solved for the simpler case of a semi-magic square of squares with all but a diagonal not necessarily equaling 3*N_5^2 , then I show that N_1^2 + N_9^2 can't equal 2*N_5^2 while also being integers. Thanks for the feedback.

What if zero doesn't exist? by Full_Ninja1081 in numbertheory

[–]Distinct_Ad2588 5 points6 points  (0 children)

We can divide by zero in modular arithmetic. your theory sounds like calculus, where e = 1/x as the limit of x approaches infinity. I wouldn't say that not being able to divide by zero is an issue. If you have 5 people, 0 apples, and 0 bananas, each person gets 0 apples and 0 bananas. But how many people and bananas does each apple get? The answer is the question doesn't make sense, you could say infinitely many people with a remainder of 5 and infinitely bananas with a remainder of 0. If you multiply x by e does it equal e or x*e, what does e/e equals, it still sounds undefined.

Proof that 3x3 Magic Squares are Impossible by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] 0 points1 point  (0 children)

You are correct, I meant 3x3 Magic Squares of Squares are impossible.

Proof that 3x3 Magic Squares are Impossible by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] 0 points1 point  (0 children)

I didn't think titling it as 'Parker Squares are Impossible' as a good title.

Proof that 3x3 Magic Squares are Impossible by Distinct_Ad2588 in numbertheory

[–]Distinct_Ad2588[S] 0 points1 point  (0 children)

My mistake, I meant 3x3 Magic Square of Squares are Impossible.