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May 31st, 2018, 01:20 AM  #1 
Senior Member Joined: Dec 2006 Posts: 167 Thanks: 3  Equation a^2+b = n(a+b^2)
Prove or disprove: For every $n\in\mathbb{N}^+$, there are $a,\,b\in\mathbb{N}^+$ such that $a^2+b=n(a+b^2)$ 
May 31st, 2018, 03:23 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 882 
Well...in case you're interested, there are 40 cases where n is an integer>0, a>b, and a<100. (a,b,n) 1: 5,2,3 2: 5,3,2 3: 10,4,4 ... 38: 84,12,31 39: 88,8,51 40: 89,17,21 
May 31st, 2018, 06:05 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,286 Thanks: 1681 
If a can exceed 100, can you find a solution when n = 33?

May 31st, 2018, 07:39 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 882 
That'll result in a^2  33a + b  33b^2 = 0 a = {33 + SQRT[33^2  4(b  33b^2)]} / 2 which has no integer solution...no? 
June 1st, 2018, 07:12 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,286 Thanks: 1681 
At present, I don't know whether it has a solution in positive integers. If $b$ can be 1, ($a$, $b$, $n$) = ($n$ + 1, 1, $n$) is a solution for any $n$. 
June 1st, 2018, 09:39 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 882 
Well, none up to b = 2000000 Closest is b = 1245121 : then a = 7152659.0000002796..... 
June 1st, 2018, 09:49 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 882 
Well, decided to go up to b = 10 million; found 2 of the mudders: b = 4,538,727 : a = 26,073,018 b = 6,384,248 : a = 36,674,729 However, I'm sure that "discovery!" will not bring down the price of groceries 

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