Replies to This Discussion

Permalink Reply by Keith Brian Johnson on February 18, 2013 at 1:57am

This doesn't really answer your question, but we can suppose f(m)+n=f(n)+m, making the product a perfect square for all m, n. Then n-m=f(n)-f(m), and then we can take

f(k)=g(x1,x2,x3,...,k)+k for any natural number k and variables x1,x2,x3,..., as long as g is defined at k. So, for example, any equation of a line in slope-intercept form will do, if k is the y-intercept. If, say, f(k)=3x+k, then f(m)=3x+m and f(n)=3x+n, and (3x+m+n)(3x+n+m) is a perfect square.

 

But you're asking whether such functions provide the only solutions. Hmm. That could take a while.

Permalink Reply by Luara on February 18, 2013 at 5:27am

If you guess that f(m)+n = f(n)+m for all natural no's m and n, that means

f(m)-m=f(n)-n for all m,n, i.e. f(n)-n is a constant.  Say f(n)-n=k, that would mean f(n)=n+k. 

But is that guess true, does f(n)-n have to be a constant?  What are the possible choices for f, as a function of (one) natural number to the naturals?

You could make f a function of other variables too, I guess.  I doubt that helps in determining f as a function of one variable, though. 

Permalink Reply by Luara on February 25, 2013 at 10:17am

So, would people like to see my answer to this puzzle?  or are you still working on it and would like more time?  are you interested in the answer or not?