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- TL;DR Summary
- For the function f(m,n)=m^2n^2/[(m+n)(m-n)], are there two distinct pairs of integers (m1,n1), (m2,n2) such that f(m1,n1)=f(m2,n2)?

Hello all,

This is a problem of a different flavour from my usual shenanigans. I'm looking at a function

$$f(m,n)=\frac{m^2n^2}{(m+n)(m-n)}$$

and am trying to determine if there are any two pairs of values ##(m_1,n_1)## and ##(m_2,n_2)## which evaluate to the same result. Assume that ##m_i>n_i## for ##i=1,2##.

I've made a massive table in Google sheets that calculates the fraction (well, reduced numerator and denominator) for values of ##m,n## between ##1## and ##99## (!). I don't see any matches but I'm not convinced there can't be any.

So I've tried to prove the following: If ##f(m_1,n_1)=f(m_2,n_2)## then ##m_1=m_2## and ##n_1=n_2##.

It seems like the best approach is using modular arithmetic (I'm most familiar with dealing with squares mod 10). I would write

$$m_2^2=\frac{m_1^2n_1^2(m_2^2-n_2^2)}{n_2^2(m_1^2-n_1^2)}$$

and then maybe try to show that the RHS is not the square of an integer unless ##m_1=km_2## and ##n_1=kn_2##.

Alternatively, I can use the fact that if ##m_1,n_1## are relatively prime, then ##m_1\pm n_1## is relatively prime to both of them. But then I get something which depends on the prime factors of ##m_1## and ##n_1##. I don't really know how to proceed - this is probably something number theory can help with and is outside my domain of expertise.

Any suggestions?

Thanks,

QM

EDIT 1: I think I see another way forward. Given ##m_1,n_1##, we have ##f(m_1,n_1)=N/D## for integers ##N,D##; then ##f(Dm_1,Dn_1)=DN## which is an integer. Maybe I can show that these integers are all distinct...

EDIT ##e##: I don't think that's the best way forward anymore. I'm going to try making my argument clear with a simplified case.

EDIT ##\pi##: Okay, here's the best I've got.

I want to prove the following claim:

If ##a,b,c,d## are integers and ##a^{-1}-b^{-1}=c^{-1}-d^{-1}##, then ##a=c## and ##b=d##.

Now, we can rearrange to get ##d^{-1}=c^{-1}+b^{-1}-a^{-1}##, so

$$d=\frac{abc}{ab+ac-bc}.$$

If ##a=c##, then ##ab-bc=0## and ##d=b##. Otherwise assume ##a\neq c##.

Also assume that ##\mathrm{gcd}(a,b,c)=1##. Then ##ab+ac-bc## is not a multiple of ##ab,c## (write it as ##ab+(a-b)c##), ##ac,b## (##ac+(a-c)b##), or ##bc,a## (##a(b+c)-bc##). Therefore ##\frac{abc}{ab+ac-bc}## can only be an integer if ##ab+ac-bc## takes some factors of ##a,b,c## but not the ones needed to form a multiple of one of ##a,b,c,ab,ac,bc##.

I believe I can conclusively prove the lemma for primes and squares of primes, but I need to have a think about how to get from there to arbitrary squares ##a,b,c,d##. My rigourous maths is a bit rusty.

EDIT ##\delta_\mathrm{feigenbaum}##: I don't think my primes argument holds water.

Suppose ##a,b,c,d## are primes and ##a^{-1}-b^{-1}=c^{-1}-d^{-1}##. Then

$$d=\frac{abc}{ab+ac-bc}.$$

Therefore ##ab+ac-bc## is one of ##1,a,b,c,ab,ac,bc,abc##.

If ##ab+ac-bc=ab## then ##a=b## (trivial case).

If ##ab+ac-bc=ac## then ##a=c## (what we started with).

If ##ab+ac-bc=bc## then ##2bc=a(b+c)##. But then either ##a=2## (so ##bc=b+c## and therefore ##a=b=c=2##), ##a=b## (trivial case), or ##a=c## (what we started with).

If ##ab+ac-bc=a## then ##a(b+c-1)=bc##. Either ##a=b## or ##a=c##.

If ##ab+ac-bc=b## then ##b(a-c-1)=ac##. Then ##b=a## (if ##b=c## then ##-c^2=c## which is impossible).

If ##ab+ac-bc=c## then ##c(a-b-1)=-ab##. Then ##c=a## or ##c=b##. (?)

If I can get this lemma for primes, I can get it for squares of primes using the difference of two squares formula. But then I need to generalize it to arbitrary integer squares.

I'm stuck (and also out of clever constants whose integer truncations are consecutive integers). Any help would be appreciated.

This is a problem of a different flavour from my usual shenanigans. I'm looking at a function

$$f(m,n)=\frac{m^2n^2}{(m+n)(m-n)}$$

and am trying to determine if there are any two pairs of values ##(m_1,n_1)## and ##(m_2,n_2)## which evaluate to the same result. Assume that ##m_i>n_i## for ##i=1,2##.

I've made a massive table in Google sheets that calculates the fraction (well, reduced numerator and denominator) for values of ##m,n## between ##1## and ##99## (!). I don't see any matches but I'm not convinced there can't be any.

So I've tried to prove the following: If ##f(m_1,n_1)=f(m_2,n_2)## then ##m_1=m_2## and ##n_1=n_2##.

It seems like the best approach is using modular arithmetic (I'm most familiar with dealing with squares mod 10). I would write

$$m_2^2=\frac{m_1^2n_1^2(m_2^2-n_2^2)}{n_2^2(m_1^2-n_1^2)}$$

and then maybe try to show that the RHS is not the square of an integer unless ##m_1=km_2## and ##n_1=kn_2##.

Alternatively, I can use the fact that if ##m_1,n_1## are relatively prime, then ##m_1\pm n_1## is relatively prime to both of them. But then I get something which depends on the prime factors of ##m_1## and ##n_1##. I don't really know how to proceed - this is probably something number theory can help with and is outside my domain of expertise.

Any suggestions?

Thanks,

QM

EDIT 1: I think I see another way forward. Given ##m_1,n_1##, we have ##f(m_1,n_1)=N/D## for integers ##N,D##; then ##f(Dm_1,Dn_1)=DN## which is an integer. Maybe I can show that these integers are all distinct...

EDIT ##e##: I don't think that's the best way forward anymore. I'm going to try making my argument clear with a simplified case.

EDIT ##\pi##: Okay, here's the best I've got.

I want to prove the following claim:

If ##a,b,c,d## are integers and ##a^{-1}-b^{-1}=c^{-1}-d^{-1}##, then ##a=c## and ##b=d##.

Now, we can rearrange to get ##d^{-1}=c^{-1}+b^{-1}-a^{-1}##, so

$$d=\frac{abc}{ab+ac-bc}.$$

If ##a=c##, then ##ab-bc=0## and ##d=b##. Otherwise assume ##a\neq c##.

Also assume that ##\mathrm{gcd}(a,b,c)=1##. Then ##ab+ac-bc## is not a multiple of ##ab,c## (write it as ##ab+(a-b)c##), ##ac,b## (##ac+(a-c)b##), or ##bc,a## (##a(b+c)-bc##). Therefore ##\frac{abc}{ab+ac-bc}## can only be an integer if ##ab+ac-bc## takes some factors of ##a,b,c## but not the ones needed to form a multiple of one of ##a,b,c,ab,ac,bc##.

I believe I can conclusively prove the lemma for primes and squares of primes, but I need to have a think about how to get from there to arbitrary squares ##a,b,c,d##. My rigourous maths is a bit rusty.

EDIT ##\delta_\mathrm{feigenbaum}##: I don't think my primes argument holds water.

Suppose ##a,b,c,d## are primes and ##a^{-1}-b^{-1}=c^{-1}-d^{-1}##. Then

$$d=\frac{abc}{ab+ac-bc}.$$

Therefore ##ab+ac-bc## is one of ##1,a,b,c,ab,ac,bc,abc##.

If ##ab+ac-bc=ab## then ##a=b## (trivial case).

If ##ab+ac-bc=ac## then ##a=c## (what we started with).

If ##ab+ac-bc=bc## then ##2bc=a(b+c)##. But then either ##a=2## (so ##bc=b+c## and therefore ##a=b=c=2##), ##a=b## (trivial case), or ##a=c## (what we started with).

If ##ab+ac-bc=a## then ##a(b+c-1)=bc##. Either ##a=b## or ##a=c##.

If ##ab+ac-bc=b## then ##b(a-c-1)=ac##. Then ##b=a## (if ##b=c## then ##-c^2=c## which is impossible).

If ##ab+ac-bc=c## then ##c(a-b-1)=-ab##. Then ##c=a## or ##c=b##. (?)

If I can get this lemma for primes, I can get it for squares of primes using the difference of two squares formula. But then I need to generalize it to arbitrary integer squares.

I'm stuck (and also out of clever constants whose integer truncations are consecutive integers). Any help would be appreciated.

Last edited: