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# Fun math puzzle

Here's a math puzzle for you:

Suppose f is a function from the natural numbers (1,2,3...) to the natural numbers, and (f(m)+n)(f(n)+m) is a square for all m,n. What functions f have this property?

It's easy to guess what f has to be, but the puzzle is to prove that the possibilities for f are limited to your guess - or show what other possibilities there are.

If you happen to see the solution online somewhere, please don't post it!  I'm curious about what people can come up with on their own.  I solved it (I think!).

It's a great puzzle because there are so many possible ways to tackle the problem.  And likely, various possible proofs.

Views: 34

### Replies to This Discussion

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.

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.

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?

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