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Algebra and number theory in junior high school mathematics competition
I thought about it for a long time. I think the first question is more difficult. Of course, I think you forgot to put the brackets.

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.