Consider all possible pairs (M, N) of positive integers that satisfy this equation:
M2 = N(M+N) - 1
• If this equation has a finite number of solutions, then list all of them.
• If the total number of solutions is infinite, then find the general forms of M and N with valid reasoning.
Note: Extra credit wii be given for a semi-analytic (hand calculator and p&p) solution.
Replacing M,N, with x,y, x^2-xy-y^2=-1, I attempt a p+p-ish solution.
It looks very much as though the x/y values are likely to converge to something, so start by looking at small values:
x y
1 1
3 2
8 5
Average the ratios x(n)/x(n-1) and y(n)/y(n-1)
From which ((2.666)+(2.5)/2)=2.58333. Call this average d1.
Multiply x and y by d1:
8*d1 5*d1
20.666 12.916
Take the nearest integers:
21 13
Rinse and repeat:
((2.625)+(2.6)/2) 2.6125 say, d2
21*d2 13*d2
54.8625 33.9625
Nearest integers
55 34 2.619047619 2.615384615
etc, so
2.617216117
144 89 2.618181818 2.617647059
2.617914439
377 233 2.618055556 2.617977528
2.618016542
987 610 2.618037135 2.618025751
2.618031443
2584 1597 2.618034448 2.618032787
2.618033617
6765 4181 2.618034056 2.618033813
This looks like phi, or (1+sqrt(5))/2
Moreover, the ratio of the {x,y} terms is the same, e.g:
6785/4181 = 1.618034
We next observe that, starting from any given term {x(n),y(n)} say {8,5}, x(n+1) =(2xn+yn), and y(n+1)= (xn+yn), e.g.
2*8+5 8+5
21 13
To check this, let (2a+b)/(a+b) = (sqrt(5)+1)/2
Then b = 1/2 (sqrt(5)-1)a as expected.
Thus, for all n:
x = ((1/2(sqrt(5)+1))^(2n) - (1/2(sqrt(5)-1))^(2n))/sqrt(5)
y = ((1/2(sqrt(5)+1))^(2n-1)+(1/2(sqrt(5)-1))^(2n-1))/sqrt(5)
where the first terms reflect the above findings, and the second terms are the conjugates of the first ones, to cancel out the unwanted decimal places.
So there is an infinity of solutions.
Edited on February 22, 2023, 12:39 am
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Posted by broll
on 2023-02-22 00:18:31 |
Possible Solution
