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I am looking into which quadratics factor with the following property: $x^2±ax±b$ factors e.g. all the following factor for $a=5, b=6$

$x^2+5x+6=(x+2)(x+3)$

$x^2-5x+6=(x-2)(x-3)$

$x^2+5x-6=(x+6)(x-1)$

$x^2-5x-6=(x-6)(x+1)$

I have found $x^2+10x+24, x^2+20x+96, ..., x^2+5 \cdot 2^{n-1}x+6\cdot 4^{n-1}$ all have this property. I am looking for others or is this the only family of MONIC quadratics that have this property?

Bob
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1 Answers1

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You're looking for $a^2\pm 4b$ to be perfect squares, say$$a^2\pm 4b=(c\pm d)^2=c^2+d^2\pm 2cd\iff a^2=c^2+d^2,\,b=cd/2.$$So this reduces to Pythagorean triples viz.$$c=2klm,\,d=(k^2-l^2)m,\,a=(k^2+l^2)m,\,b=klm^2(k^2-l^2).$$In your example $k=2,\,l=m=1$.

J.G.
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  • Is there any way to make b a square number too? – Bob Feb 06 '21 at 23:31
  • Could this be generalised for quadratics of the form ax^2+bx+c? – Bob Feb 07 '21 at 01:52
  • @Bob You'd have to solve $b^2\pm4ac=(d\pm e)^2$. I might add that later, but I'm not sure about the square-$b$ idea. – J.G. Feb 07 '21 at 07:18
  • @J.G. I replaced c with ac and changed c and d to d and e. But I don't have an equation for a on it's own. – Bob Feb 07 '21 at 11:45
  • @Bob No, you wouldn't: only $ac$ matters, so you're free to break its value into any two-factor factorization. – J.G. Feb 07 '21 at 11:59
  • @J.G. See I did that but sometimes a,b,c are not co-prime so it creates repeated equations. Anyway around this without manually checking them? – Bob Feb 07 '21 at 13:50
  • @Bob Not if $ac$ isn't coprime to $b$. – J.G. Feb 07 '21 at 13:55
  • see also: https://math.stackexchange.com/questions/3986029/find-the-integer-pair-a-b-that-satisfies-this –  Dec 31 '21 at 05:25