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May 31st, 2018, 02:20 AM   #1
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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 $a^2+b=n(a+b^2)$
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May 31st, 2018, 04:23 PM   #2
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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
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May 31st, 2018, 07:05 PM   #3
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If a can exceed 100, can you find a solution when n = 33?
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May 31st, 2018, 08:39 PM   #4
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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?
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June 1st, 2018, 08:12 AM   #5
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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 $n$.
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June 1st, 2018, 10:39 AM   #6
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Well, none up to b = 2000000

Closest is b = 1245121 : then a = 7152659.0000002796.....
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June 1st, 2018, 10:49 AM   #7
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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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