My Math Forum Equation a^2+b = n(a+b^2)
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 May 31st, 2018, 02:20 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, 04:23 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,984 Thanks: 995 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, 07:05 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,307 Thanks: 1976 If a can exceed 100, can you find a solution when n = 33?
 May 31st, 2018, 08:39 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,984 Thanks: 995 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, 08:12 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,307 Thanks: 1976 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, 10:39 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,984 Thanks: 995 Well, none up to b = 2000000 Closest is b = 1245121 : then a = 7152659.0000002796.....
 June 1st, 2018, 10:49 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,984 Thanks: 995 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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