Find the largest possible integer pair (m, n) which satisfy (m2-mn –n2)2=1. where m, n?

Answer 1

I am going with answer #1 unless you meant the 2 as an exponent in which case you would have (m^2-mn-n^2) ^2 =1. Take square roots to get m^2-mn-n^2 = +/- 1. Then try a few.

I'd guess that there infinitely many, like 4181, 2584 and 2584, 1597 and 1597, 987. In fact these form rational approximations to the golden ratio......... Oh yeah, the Fibonacci numbers! Isn't there some property like (Fn)^2-Fn*F(n-1)-F(n-1)^2 = +/- 1 ? Anyway, it is unbounded. Trick question.

Answer 2

so...we are to solve this diophantine equation:

(m^2 - mn - n^2)^2 = 1

it's an easy one:

m^2 - mn - n^2 = +1 or m^2 - mn - n^2 = -1

so we start with the first case:

m^2 - mn - n^2 = 1
solve for m:
m^2 - nm - (n^2 + 1) = 0

so the discriminant is: n^2 + 4(n^2 + 1)
since we are trying to find integer soloutions,the discriminant must be a mean square,so the square root is an integer:

n^2 + 4(n^2 + 1) = k^2
n^2 +4n^2 + 4 = k^2
5n^2 = k^2 - 4
5n^2 = (k-2)(k+2) =>
1) k - 2 = 5 => k = 7
k + 2 = n^2 => 9 = n^2 => n = 3 or -3
put in the main equation: m^2 - 3m - 10 = 0
=> (m-5)(m+3) = 0 => m = 5 or m = -3
so the first and the second soloutions are: (5 , 3) , (-3 , 3)

m^2 + 3m -10 = 0
(m+5)(m - 2) = 0 => m = -5 or m = 2

so we find the next two: (-5 , -3) , (2 , -3)

2) k + 2 = 5 => k = 3
k - 2 = n^2 => 1 = n^2 => n = 1 or n = -1
put in the equation: m^2 - m - 2 = 0
=> (m-2)(m+1) => m = 2 or m = -1

here are the fifth and the sixth : (2 , 1) , (-1 ,1)

m^2 + m -2 = 0
=> (m-1)(m+2) = 0 => m = 1 or m = -2
so 7th and 8th are: (1 , -1) , (-2 , 1)

now we continue with the case two:

m^2 - mn - n^2 = -1
we solve it for m:
m^2 - nm - (n^2 - 1) = 0
and again:

n^2 + 4(n^2 - 1) = k^2
n^2 + 4n^2 - 4 = k^2
5n^2 - 4 = k^2
4n^2 - 4 +n^2 = k^2
4(n^2 - 1) = k^2 - n^2
=> 4(n-1)(n+1) = (k-n)(k+n)

1) k - n = 4(n-1) => k = 5n - 4
k + n = n + 1 => k = 1 => 1 = 5n - 4 => n = 1
in main equation:

m^2 - m = 0 => m(m -1) = 0 => m = 0 or m = 1
9th and 10th go to (0,1) , (1,1)

2) k + n = 4(n-1) => k = 3n - 4
k - n = n + 1 => k = 2n + 1 => 3n - 4 = 2n + 1 => n = 5

m^2 - 5m - 24 = 0 => (m + 3)(m - 8) => m = -3 or m = 8
so 11th and 12th are: (-3 , 5) and (8, 5)

3)k + n = n -1 = > k = -1
k - n = 4(n+1) = > k = 5n + 4 => -1 = 5n + 4 =< n = -1

m^2 + m = 0 => m = 0 or m = -1 so
( -1 , -1) and ( 0 , -1) are our 13th and 14th

4)k + n =4(n + 1) => k = 3n + 4
k - n = n - 1 => k = 2n - 1 => 3n + 4 = 2n - 1 => n = -5

m^2 + 5m - 24 = 0 => (m + 8)(m - 3) = 0
=> m = -8 or m = 3
so our last soloutions are
15th and 16th soloutions: (-8 , -5) , ( 3 , -5)

so the largest possible integer pair is (8 , 5)

Answer 3

Now m2 –mn –n2 = s, where s=-1 or s=+1; thence m2 –mn –n2 –s =0 (eq1), hence
m=(n +c*r)/2, where integer r= sqrt(5n^2 +4s), c=-1 or c=+1;
now r>n thus c=1 as m is supposed to be greater than 0; so m=(n+r)/2;
here come 2 expressions for m:
m=(n+r1)/2, r1=sqrt(5n^2-4) and
m=(n+r2)/2, r2=sqrt(5n^2+4)
I omit the proof, but I state:
The number of possible integer pairs (m, n) is not limited;
Thus no largest pair exists!

Answer 4

First, i got here across a relationship that gets from one exponent to the subsequent. (a+b?5)/2 * (a million+?5)/2 = ((a+5b)/2 + (a+b)/2 ?5)/2 n | a | b | a/b a million a million a million a million--> (a million+5)/2=3 and (a million+a million)/2=3 2 3 a million 3-->(3+5)/2=4 and (3+a million)/2=2 3 4 2 2-->(4+5*2)/2=7 and (4+2)/2=3 4 7 3 2.33-->(7+5*3)/2=eleven and (7+3)/2=5 5 eleven 5 2.2 6 18 8 2.25 7 29 13 2.230769231 8 40 seven 21 2.238095238 observe that the ratio strategies ?5 so... (29 + 13?5)/2 = 13(29/13 + ?5)/2?13(29/13 + 29/13)/2=29 21(40 seven/21 + ?5)/2?21(40 seven/21+40 seven/21)/2=40 seven etc. with the aid of fact the exponent gets bigger, the ratio gets closer to ?5 and the approximation gets closer to being wonderful for that reason the value gets closer to an integer.

Answer 5

I'm pretty sure there are no such pairs.
If you divide both sides by 2, then m^2-mn-n^2 = .5, and if both m and n are integers, their squares, products and sums are also integers, and .5 is not an integer.