Hearted Youtube comments on Mathologer (@Mathologer) channel.
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Hey Mathologer,
My solution to the Strand Problem has a much more intuitive way to find the x^2+x = 2y^2 identity.
It helps a lot to instead write in this way:
x(x+1)/2 = y^2
x is the largest number house, and 1 is the 1st house. The sum of all house numbers is the average of the highest house x and 1, so (x+1)/2, multiplied by the number of houses there are, which is x.
Ergo, sum of all houses is x(x+1)/2.
If the sum of all houses is equal to a perfect square, y^2, then the occupant lives in the house y.
After you reach this point in the puzzle, you can solve with wolfram to find all solutions.
(1,1) (6,8) (35,49) (204,288) (1189,1681)
Wolfram gives an equivalent formula to the one you reach at end of video. Would have done it myself but too lazy 🙂
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I was introduced to the Steinbach ratios 8 years ago. When I am working with the heptagon ratios, I prefer using the three side lengths of a heptagon with the smallest side negated:
a=2sin(τ/7), b=2sin(2τ/7), c=2sin(4τ/7)
Using these values, any polynomial f(x,y,z) such that f(a,b,c)=0 also has f(b,c,a)=0 and f(c,a,b)=0
It also has the property that ab+ca+bc=a+b+c+abc=a²+b²+c²-7=0
Also, to expand on your formulas at 43:00, A(m,n)=C(m,n+1), B(m,n)=C(m+1,n),
and C(m,n)=(1*(r/1)^m*(s/1)^n+r²*(-s/r)^m*(1/r)^n+s²*(1/s)^m*(-r/s)^n)/(1+r²+s²)
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