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

Question

where m, n
are positive integers, less than 13289.

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 ? So trace the Fibonacci sequence out to 13k: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, .... so then pick m=10946 and n=6765

Answer 2

so...we are to unravel this diophantine equation: (m^2 - mn - n^2)^2 = a million it quite is an common one: m^2 - mn - n^2 = +a million or m^2 - mn - n^2 = -a million so we start up with the 1st case: m^2 - mn - n^2 = a million resolve for m: m^2 - nm - (n^2 + a million) = 0 so the discriminant is: n^2 + 4(n^2 + a million) for the reason that we are looking for integer soloutions,the discriminant would desire to be a imply sq.,so the sq. root is an integer: n^2 + 4(n^2 + a million) = ok^2 n^2 +4n^2 + 4 = ok^2 5n^2 = ok^2 - 4 5n^2 = (ok-2)(ok+2) => a million) ok - 2 = 5 => ok = 7 ok + 2 = n^2 => 9 = n^2 => n = 3 or -3 put in the significant equation: m^2 - 3m - 10 = 0 => (m-5)(m+3) = 0 => m = 5 or m = -3 so the 1st and the 2nd soloutions are: (5 , 3) , (-3 , 3) m^2 + 3m -10 = 0 (m+5)(m - 2) = 0 => m = -5 or m = 2 so we come across the subsequent 2: (-5 , -3) , (2 , -3) 2) ok + 2 = 5 => ok = 3 ok - 2 = n^2 => a million = n^2 => n = a million or n = -a million put in the equation: m^2 - m - 2 = 0 => (m-2)(m+a million) => m = 2 or m = -a million indexed below are the 5th and the 6th : (2 , a million) , (-a million ,a million) m^2 + m -2 = 0 => (m-a million)(m+2) = 0 => m = a million or m = -2 so seventh and eighth are: (a million , -a million) , (-2 , a million) now we proceed with the case 2: m^2 - mn - n^2 = -a million we resolve it for m: m^2 - nm - (n^2 - a million) = 0 and lower back: n^2 + 4(n^2 - a million) = ok^2 n^2 + 4n^2 - 4 = ok^2 5n^2 - 4 = ok^2 4n^2 - 4 +n^2 = ok^2 4(n^2 - a million) = ok^2 - n^2 => 4(n-a million)(n+a million) = (ok-n)(ok+n) a million) ok - n = 4(n-a million) => ok = 5n - 4 ok + n = n + a million => ok = a million => a million = 5n - 4 => n = a million in significant equation: m^2 - m = 0 => m(m -a million) = 0 => m = 0 or m = a million ninth and tenth bypass to (0,a million) , (a million,a million) 2) ok + n = 4(n-a million) => ok = 3n - 4 ok - n = n + a million => ok = 2n + a million => 3n - 4 = 2n + a million => n = 5 m^2 - 5m - 24 = 0 => (m + 3)(m - 8) => m = -3 or m = 8 so eleventh and twelfth are: (-3 , 5) and (8, 5) 3)ok + n = n -a million = > ok = -a million ok - n = 4(n+a million) = > ok = 5n + 4 => -a million = 5n + 4 =< n = -a million m^2 + m = 0 => m = 0 or m = -a million so ( -a million , -a million) and ( 0 , -a million) are our thirteenth and 14th 4)ok + n =4(n + a million) => ok = 3n + 4 ok - n = n - a million => ok = 2n - a million => 3n + 4 = 2n - a million => n = -5 m^2 + 5m - 24 = 0 => (m + 8)(m - 3) = 0 => m = -8 or m = 3 so our final soloutions are fifteenth and sixteenth soloutions: (-8 , -5) , ( 3 , -5) so the main important a threat integer pair is (8 , 5)

Answer 3

following m and n, are those 2s or squares? same question for the outside of the parenthesis.

Answer 4

I don't think language like this should be allowed on this site!

Answer 5

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

Answer 6

negative zero and super negative zero