Because k is an integer, then n/(Mn) is an integer, and m|n is obtained.
As long as m=n= 1 is taken here, then k=3 is not a square number.
If not, it is n/(nm+ 1)
And then (Mn+1) | n 2,
And (mn+ 1, n)= 1, when m and n are positive integers.
This cannot be true at the same time.
So the original question should be (m 2+n 2)/(nm+ 1).
The second question is relatively simple. Just prove that 1/m+ 1/n is an integer.
If n, m>21/m+1/n <; 1。 So m, n must be less than or equal to 2.
By enumerating, m = 1, n = 1.m = 2, and n = 2 is all solutions of 1/m+ 1/n as integers.
So k=m+ 1/m+n+ 1/n=3, or 4.
The first question, I can't answer your question in simple language, but I thought of it and submitted the answer.
I estimate the range of k by estimation, and then draw a conclusion by enumeration.
We prove that k is a square number, m=n=k= 1.
Prove:
Let t = m/n and m >；; =n，
Then t>= 1:
k
=[(m^2+n^2)/(mn)]*[ 1/( 1+ 1/(mn))]
=(t+ 1/t)*( 1- 1/(mn)+ 1/(mn)^2+.....)
=t+ 1/t-( 1/m^2+ 1/n^2)+[(m^2+n^2)/(nm)^3]*( 1/( 1+ 1/(mn)))
Let s =1/t-(1/m2+1/N2)+[(m2+N2)/(nm) 3] * (1/).
Then k = t+s.
Let's estimate S.
s & lt= 1/t-( 1/m^2+ 1/n^2)+(m^2+n^2)/(mn)^3
= 1/t-( 1/n^2+ 1/m^2)( 1- 1/(mn))
& lt 1/ ton
Get a k < t+1/t.
If n & gt=2, t & gt=2.
Then, s> =1/t-(1/n 2+1/m 2) > = 1/2-5/ 16 >； 0
So t < k
So t
If m> = n 2
There is1/n > at this time; =n/m
Then k < t+1/t <; =[t]+(n- 1)/n+n/m & lt； =[t]+ 1
[t]+ 1
At this time, k=[t]+ 1, so let [t]=p,
So m=np+g, 0
With k = (m 2+n 2)/(Mn+1), after simplification:
p(n^2-ng+ 1)=n^2-ng- 1.
p =(n ^ 2-ng- 1)/(n ^ 2-ng+ 1)< 1。
This contradicts that p is an integer.
So when n & gt=2, 1
So:
s & gt= 1/t-( 1/m^2+ 1/n^2)>； = 1/t- 1/2
Then:
t+ 1/t- 1/2 & lt； K & ltT+ 1/t, because 1
So1< = k <; 3。
So k= 1 or k=2.
When k=2.
There is 2 = (m 2+n 2)/(Mn+ 1)
Get:
2=(m-n)^2
Obviously, the above equation has no integer solution.
When k= 1:
There is Mn+ 1 = m 2+n 2 > =2mn.
So Mn
But n & gt=2. So it is impossible.
Therefore, consider n= 1, when m >; =2:
k=(m^2+ 1)/(m+ 1).
Then k < m+1/m.
s & gt= 1/m-( 1+ 1/m^2)
m+ 1/m- 1- 1/m^2<； k & ltm+ 1/m
m-3/4 & lt； = k & lt=m
So we come to the conclusion that:
k=m。
m=(m^2+ 1)/(m+ 1)
m^2+m=m^2+ 1
It is concluded that m= 1, which is different from m >: =2,
So m= 1.
At this point:
k=( 1^2+ 1^2)/( 1* 1+ 1)= 1。
So:
The solution of k = (m 2+n 2)/(Mn+1) and m, n and k are positive integers is unique;
K=m=n= 1。 And k is a square number.
I have been calling for a long time. I may have made a mistake. Please forgive me. In fact, there seems to be a way to refute this question, in the following link:
/question/3 15 132027 . html
The proof here is not complicated.