Demonstration that the sum of the squares of 3 consecutive integers cannot give an integer X squared.

Consider the sequence of integers (n-1), n, (n+1)

S = the sum of the squares of this sequence = (n-1)² + n² + (n+1)² = (n² - 2n +1) + n² + (n² + 2n +1)

S = 3n² + 2

By taking the square root of S, we obtain a non-integer X, at least by trial and error because no one to date has been able to find an integer result. We are then faced with a conjecture that must be proven exactly. To demonstrate this impossibility of obtaining such an integer, we will compare the modular properties of S and X². If a modulo of S does not equal the same modulo of X², this will prove that S ≠ X².

Let's first take X|3 = the modulo 3 of X = the remainder of the division of an integer X by 3.

Three remainders for X|3 are possible when X is integer, either 0 or 1 or 2.

Now to calculate the modulo 3 of a multiplication of a b,

we use the formula: ab|3 = (a|3 × b|3) |3

X²|3 = (X × X) |3 = (X|3 × X|3) | 3

X Modulo 3 can be 0, 1, or 2

If X|3 = 0 and X is integer then X²|3 = ( 0 × 0 ) | 3 = 0

If X|3 = 1 and X is integer then X²|3 = ( 1 × 1 ) | 3 = 1

If X|3 = 2 and X is integer then X²|3 = ( 2 × 2 )|3 = 4|3 = 1

We therefore notice that the modulo of a square X²|3 for an integer X = 0 or 1

But it is never 2.

What about S|3 = ( 3n² + 2 ) | 3 = (3n²|3 + 2|3)|3 = (0 + 2) |3 = 2

Thus, modulo 3 of S is always 2, precisely the value that modulo 3 of X² can never have.

So S ≠ X² and this proves the impossibility that the sum of the squares of any 3 consecutive integers can equal the square of an integer X.

QED

The conjecture is resolved. Note that some mathematical conjectures took centuries to resolve with proofs weighing hundreds of pages of demonstrations.

The same type of proof easily applies to a sequence of 4 integers where S|4 = 2 while X²|4 cannot be neither 2 nor 3. This methodology also eliminates sequences of 9 and 12 integers...

Demonstration that the sum of the squares of 5 consecutive integers cannot give an integer X squared.

S = n² + (n+1)² + (n+2)² + (n+3)² + (n+4)² = 5n² + 20n + 30

So let's see for the modulo 4 of this sum and of X².

X²|4 = (X × X) |4 = (X|4 × X|4) | 4

If X|4 = 0 then X is integer and X²|4 = ( 0 × 0 ) | 4 = 0

If X|4 = 1 then X is integer and X²|4 = ( 1 × 1 ) | 4 = 1

If X|4 = 2 then X is integer and X²|4 = ( 2 × 2 ) | 4 = 4|4 = 0

If X|4 = 3 then X is integer and X²|4 = ( 3 × 3 ) | 4 = 9|4 = 1

So, X²|4 can only be 0 or 1 but never 2 or 3.

S|4 = (5n² + 20n + 30) | 4

If n|4 = 0: S | 4 = (0 + 0 + 30) | 4 = 30|4 = 2

If n|4 = 1: S | 4 = (5 + 20 + 30) | 4 = 55|4 = 3

If n|4 = 2: S | 4 = (20 + 40 + 30) | 4 = 90|4 = 2

If n|4 = 3: S | 4 = (45 + 60 + 30) | 4 = 135|4 = 3

Since X²|4 is always 0 or 1 for an integer X and S|4 is always 2 or 3, it is impossible for X² = S with X integer.

So the sum of the squares of 5 consecutive integers cannot give an integer X squared.

QED

And the sum of the squares of 13 consecutive integers cannot give an integer X squared because also the modulo 4 of this sum is always 2 or 3.

This method also applies to a sequence of 6 consecutive integers because the modulo 4 of the sum of the squares is always 3.

Impossibility demonstrated so far with the properties of modulo for the sequences of 3 4 5 6 12 13.

Good news is the happy possibility of this fabulous sequence of 11 integers:

18² + 19² + 20² + 21² + 22² + 23² + 24² + 25² + 26² + 27² + 28² = 77²

And, also remarkable is this consecutive sequence of 3 squares: 10² + 11² + 12² = 13² + 14²

By the way :

3435 = 3³ + 4⁴ + 3³ + 5⁵

73939133 is prime and all its truncations

7393913

739391

73939

7393

739

7 Can be verified with this Prime calculator

Can a 3x3 magic square be constructed with nine distinct square numbers?

Still (2024), no one found such a magic square; so a conjecture of impossibility is to be considered. Can it be proved?

By Richard Lefebvre and Michel Aubé