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Find the number of integers or find all integers for which \(x^{2}+bx+c\) becomes a perfect square of an integer, where \(b,c\) are integers.I think this would be very tough as we are not given any information about \(b,c\).What do you think.?

Note by Kishan K
3 years, 3 months ago

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If \( c = \left( \frac{b}{2}\right)^2 \), then

\[ x^2+bx+c=x^2+2\cdot\frac{b}{2}\cdot x+\left( \frac{b}{2}\right)^2=\left( x + \frac{b}{2}\right)^2. \]

So, if \((b,c) \in \{ \dots ,(-6,9),(-4,4),(-2,1),(0,0),(2,1),(4,4),(6,9),\dots \},\)

then it is a perfect square for all \(x \in \mathbb{Z}\). Furthermore, if \(c\) is a perfect square, then \(x=0\) always yields a perfect square, regardless of \(b\).

I'd say those are the special cases. I wouldn't know a general way to find all perfect squares for some integers \(b,c\). Tim Vermeulen · 3 years, 3 months ago

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@Tim Vermeulen Thnx,I also think that it is very tough.But it would be very helpful if we get some general results... Kishan K · 3 years, 3 months ago

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As Tim pointed out, there the class of functions of perfect square polynomials yield infinitely many perfect square values. For all of the other cases, there are finitely many solutions.

To deal with the general case, you want to bound \(x^2 + bx +c \) strictly between 2 consecutive squares, for all but finitely many values of \(x\). The squares \( (x + \lfloor \frac{b}{2} \rfloor)^2 \) and \( (x + \lceil \frac{ b}{2} \rceil)^2 \) should work. Calvin Lin Staff · 3 years, 2 months ago

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