Let the positive integer m, n satisfy m n, and 1 m 1 n 1 1 23 then m n

1/m-1/(n+1)

n-m+1)/[m(n+1)]

23(n+1-m)=m(n+1);

23(n+1)-23m=m(n+1);

m=23(n+1)/(n+24)……1), because m is a positive integer, 23(n+1)=k(n+24);

23n+23=kn+24k;

23-k)n=24k-23……(2);

Because m23n+23;

n^2+n-23>0;The solution is n>[-1+root(93)] 2 > 1+9] 2=4, i.e., n>=5

On the other hand, k cannot be taken as an odd number, because when k is an odd number, 23-k is even and (23-k)n is even, and the right side of equation (2) is odd, so the equation cannot be true.

So k can only take 6,8,10,12,14,16,18,20,22

And easy to observe k can only take 22. In fact, 23-k can only be 1 to ensure that 24k-23 is divisible by 23-k.

For example, if k=20, 23-k=3; The 23 in the 24*3-23 on the right does not contain a factor of 3, so it is not divisible by the left.

If k=18, 23-18=5; The 24*18-23 on the right does not contain a factor of 5, so it is not divisible by the left.

In the same way, it is easy to see that nothing else is good. So k=22

So k=22n=505, m=22 m+n=527.

23-m)n=24m-23,n=(24m-23)/(23-m)=23m/(23-m)-1……Since m and n are positive integers, then 24m-23 0, then 23-m 0 can be obtained, because n is an integer, then 23m (23-m) can be obtained as an integer, because 23 is a prime number, so (23-m) must be a factor of 23 or m, if it is a factor of 23, since 23-m ≠ 23, so 23-m = 1, that is, m = 22, substituted to obtain n = 23 * 22-1 = 505, so we get the first set of solutions (m, n) = (22,505) ......If 23-m is the factor of m, let m=(23-m)k, that is, m=23k (1+k)=23-23 (1+k), k has a unique solution k=22, and substituting it gives m=22, n=505, so there is only one set of solutions that satisfy the problem: (m,n)=(22,505).

Multiply by 4 to get 4m (2n 7) 39, decompose to get (2m 2n 7)(2m 2n 7) 39, since m and n are both positive integers, so 2m 2n 7 39, 2m 2n 7 1;Or.

2m 2n 7 13, 2m 2n 7 3, respectively, solve the above two equations, you can get.

m＝10，n＝6 。

Summary. It's almost there.

If m,n is a positive integer, and m -n -7n-22 = 0, find the value of m,n.

What do you think? It's almost there.

Right away. It's okay, dear.

Good. Isn't that minus n.

How did your grouping become plus n.

I'll do the math again.

Misread. I'm counting one step.

m=10n=6

You're right to bring it in.

10 and 6 are right to bring in. You try.

It's not right.

Kiss this ** you sent me, it's obvious.

Wrong. I was right, but I was said to be at a bad level.

m=10 n=6

That's right. You're that ** inside.

The answers are all negative.

The question requirement is positive.

And this ** inside the process is obviously flickering.

I hope you can see the right answer.

n/m+(m+1)/n=4;+m, n are positive integers, solve n, m

m(n+1)-n(m+1)] n(n+1)]=m-n) [n(n+1)], because it is m, n is a positive number of rotten books, so n(n+1) is obviously a positive integer, and the sign of m-n cannot be determined, it should be determined according to the macro value of m, n

(x) is the number of Euler Hanpeng rocks.

It is the number of numbers that are coprime with x in numbers less than or equal to x.

When (m,n)=1 there is.

mn)=φm)*φn)

If (m,n) is not = 1

mn)>=m)*φn)

Usually wood has a concern (mn)= m)+ n).

And if n>=3, then (n)>=2

So if m and n are both = 3 then.

mn)>=3 max>φ(m)+φn)

So m,n at least one is 2

2n)=φ2)+φn)

2n)=1+φ(n)

Let n be given by the prime numbers p, q, .,, r divisible.

If touching and high n are odd numbers.

n)=n(1-1/p)(1-1/q)..1-1/r)(2n)=2n(1-1/p)(1-1/q)..1-1/r)(1-1/2)

No way. If n is an even number.

n)=n(1-1/p)(1-1/q)..1-1/r)(2n)=2n(1-1/p)(1-1/q)..1-1/r)(n)=n(1-1/p)(1-1/q)..

1-1/r)=2n=3,4,6

So (m,n) is. Or.