Fun math puzzle - Atheist Nexus2014-04-20T16:31:23Zhttp://www.atheistnexus.org/forum/topics/fun-math-puzzle?groupUrl=mathematics&commentId=2182797%3AComment%3A2166664&xg_source=activity&groupId=2182797%3AGroup%3A68303&feed=yes&xn_auth=noSo, would people like to see…tag:www.atheistnexus.org,2013-02-25:2182797:Comment:21699942013-02-25T16:17:37.913ZLuarahttp://www.atheistnexus.org/profile/Luara
<p>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?</p>
<p>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?</p> If you guess that f(m)+n = f(…tag:www.atheistnexus.org,2013-02-18:2182797:Comment:21666642013-02-18T11:27:55.119ZLuarahttp://www.atheistnexus.org/profile/Luara
<p>If you <em>guess</em> that f(m)+n = f(n)+m for all natural no's m and n, that means</p>
<p>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. </p>
<p>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?</p>
<p>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. </p>
<p>If you <em>guess</em> that f(m)+n = f(n)+m for all natural no's m and n, that means</p>
<p>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. </p>
<p>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?</p>
<p>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. </p> This doesn't really answer yo…tag:www.atheistnexus.org,2013-02-18:2182797:Comment:21666532013-02-18T07:57:02.279ZKeith Brian Johnsonhttp://www.atheistnexus.org/profile/KeithBrianJohnson
<p>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</p>
<p>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.</p>
<p> </p>
<p>But…</p>
<p>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</p>
<p>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.</p>
<p> </p>
<p>But you're asking whether such functions provide the only solutions. Hmm. That could take a while.</p>