Well there is no symmetry I can see and factoring all of this out looks a bit like a pain. The form looks a bit like the theorem of Pythagoras. We know a^2+b^2=c^2 if a=m^2-n^2 b=2mn c=m^2+n^2 for any m,n integers Comparing equations x=m^2-n^2 y=mn xy-14=m^2+n^2 mn(m^2-n^2)-14=m^2+n^2 (m^2-n^2)(mn-1)=14+2n^2 1. If n=odd right hand side will divide by 4 (7+n^2)*2 a) m is odd m+n, m-n and mn-1 will divide by 2 each so left hand divides by 8 => 7+n^2 is divisible by 4. If n=2k+1 then 7+n^2=4k^2+4k+8 is always divisible by 4 so this is worth considering (*) b) m is even so m+n, m-n and mn-1 are all odd so we also have no solution If n is even right hand side will divide by 2 only a) m=odd m+n, m-n and mn-1 are all odd and their product is also odd => no solution b) m=even then m+n ,m-n are even so Left hand divides by 4 so we need 7+n^2 to be divisible by 4 which is impossible as n is even (*) Continuation m=2j+1 and n=2k+1 (mn-1)(m^2-n^2)=14+2n^2 ((2j+1)(2k+1)-1)2(j+k)2(j-k)=8k^2+8k+16 8(2jk+j+k)(j+k)(j-k)=8k^2+8k+16 (2jk+j+k)(j+k)(j-k)=k^2+k+2 I don't see any easy way to factor so I will try some values for k. Since j,k are integers the product on the left is an integer product but equation k^2+k+2=0 does not have integer solutions so there are no easy conditions to be fulfilled by j and k, maybe just get polynomials in j and k and do long division. 2j^3k-2jk^3+j^3-jk^2+j^2k-k^3=k^2+k+2 -k^3(1+2j)-k^2*j+k(2j^3+j^2)+j^3=-k*(1+2j)*(k^2+k+2)+k^2(1+2j)+2k(1+2j)-k^2j+k(2j^3+j^2)+j^3= (-k)(1+2j)(k^2+k+2)+(k^2+k+2-k-2)+k(2j^3+j^2+4j+2)+j^3=(k^2+k+2)(1-k-2jk)+k(2j^3+j^2+4j+1)+j^3-2 For the relation to gold we need the following conditions 1-k-2jk=1(1) K(2j^3+j^2+4j+1)+J^3-2=0 (2) (1) =>k(1+2j)=0 so k =0 (2) And k=0 => J^3-2=0 no integer solutions I am curious about your solution
All solutions for x: x=14, x=-14, x=0, x=-6, x=-8, x=6, x=8
Well there is no symmetry I can see and factoring all of this out looks a bit like a pain.
The form looks a bit like the theorem of Pythagoras. We know a^2+b^2=c^2 if
a=m^2-n^2
b=2mn
c=m^2+n^2 for any m,n integers
Comparing equations
x=m^2-n^2
y=mn
xy-14=m^2+n^2
mn(m^2-n^2)-14=m^2+n^2
(m^2-n^2)(mn-1)=14+2n^2
1.
If n=odd right hand side will divide by 4 (7+n^2)*2
a) m is odd m+n, m-n and mn-1 will divide by 2 each so left hand divides by 8 => 7+n^2 is divisible by 4. If n=2k+1 then 7+n^2=4k^2+4k+8 is always divisible by 4 so this is worth considering (*)
b) m is even so m+n, m-n and mn-1 are all odd so we also have no solution
If n is even right hand side will divide by 2 only
a) m=odd m+n, m-n and mn-1 are all odd and their product is also odd => no solution
b) m=even then m+n ,m-n are even so Left hand divides by 4 so we need 7+n^2 to be divisible by 4 which is impossible as n is even
(*) Continuation m=2j+1 and n=2k+1
(mn-1)(m^2-n^2)=14+2n^2
((2j+1)(2k+1)-1)2(j+k)2(j-k)=8k^2+8k+16
8(2jk+j+k)(j+k)(j-k)=8k^2+8k+16
(2jk+j+k)(j+k)(j-k)=k^2+k+2
I don't see any easy way to factor so I will try some values for k.
Since j,k are integers the product on the left is an integer product but equation k^2+k+2=0 does not have integer solutions so there are no easy conditions to be fulfilled by j and k, maybe just get polynomials in j and k and do long division.
2j^3k-2jk^3+j^3-jk^2+j^2k-k^3=k^2+k+2
-k^3(1+2j)-k^2*j+k(2j^3+j^2)+j^3=-k*(1+2j)*(k^2+k+2)+k^2(1+2j)+2k(1+2j)-k^2j+k(2j^3+j^2)+j^3= (-k)(1+2j)(k^2+k+2)+(k^2+k+2-k-2)+k(2j^3+j^2+4j+2)+j^3=(k^2+k+2)(1-k-2jk)+k(2j^3+j^2+4j+1)+j^3-2
For the relation to gold we need the following conditions
1-k-2jk=1(1)
K(2j^3+j^2+4j+1)+J^3-2=0 (2)
(1) =>k(1+2j)=0 so k =0
(2) And k=0 => J^3-2=0 no integer solutions
I am curious about your solution
The solution he presented in the video is elegant.
By checking, the answers presented are very much correct.