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Start with Perfect Square To Divisibility By 56
1:
3N+1 = a^2, 4N+1=b^2
N=(a^2-1)/3= (b^2-1)/4
4a^2-4=3b^2-3
4a^2=3b^2+1 [1]
2:, from 4a^2=3b^2+1
Since a is even, b is odd = (2n-1)
3* (2n-1)^2+1 = 12n^2-12n+4
Rearranging,
4(3n^2-3n+1) =4(n^3-(n-1)^3), consecutive cubes.
But 4(n^3-(n-1)^3) = 4a^2, so a^2 is a difference of consecutive cubes. [2]
3: let x=n-1
(x+1)^3-x^3=a^2
I 3x^2 +3x + 1 = a^2
Note that 3x^2 +3x +1 = 3 (x+1/2)^2+1/4
II 3(2x+1)^2+1 = (2a)^2 Multiplying by 4
III 3(2x+1)^2 = (2a+1)(2a-1) Subtracting 1 from both sides and factoring.
IV Now 3 divides (2a+1), or 3 divides (2a-1); let (2x+1)=yz; so:
V EITHER 3y^2z^2=(2a+1)(2a-1); and {(2a+1)=y^2, (2a-1)=3z^2}; then y^2-2 =3z^2, but this has no solutions.
VI OR {(2a+1)=3y^2, (2a-1)=z^2} in which case 3y^2-2 = z^2 [3]
4: from 3:III, we could equally have had 3(2n-1)^2 = 3y^2z^2, when n=(yz+1)/2
N=n^2-n=1/4(yz-1)(yz+1); 4N+1=(yz)^2=b^2=(2n-1)^2
Since 3y^2-z^2=2 and y^2z^2-1=4N, and further z= (3y^2-2)^(1/2), with z^2 = (3y^2-2)
We can write 4N = 3y^4-2y^2-1 [4]
5: from 3N+1 = a^2 (given)
3N+1=(n^3-(n-1)^3)
N=n^2-n. [5]
Putting 4 and 5 together:
4(n^2-n) = 3y^4-2y^2-1
Since a triangular number is 1/2(n(n-1)), mutatis mutandis, we have now integer (i.e. y) and triangular number (i.e. N/2) such that N=3y^4-2y^2-1, with y in {1,3,11,41,153..} A001835 in Sloane.
Given that N was divisible by 56 in Perfect Square To Divisibility By 56, it must follow at once that N/2 is divisible by 28, as can readily be shown using the recurrence method for the equation 3y^2-2 = z^2. |
