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Algebra

Let \(A\) be the set of all integers \(n\) of the form \(n= a^2 + 4ab+ b^2\) where \(a\) and \(b\) are integers.
a) Show that if \(x\) and \(y\) are in \(A\), then \(xy\) is in \(A\).
b) Show that 11 is not in \(A\).
c) Show that the equation \(x^2 + 4xy + y^2 = 1\) has infinitely many integer solutions.

Note by Bulbuul Dev
6 months, 2 weeks ago

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c) \(x^2+4xy+y^2=(x+2y)^2-3y^2\) which is of the form \(a^2-3b^2\). Since \(3\) is not a perfect square by Pell's equation there are infinitely many solutions.

Brilliant Member - 6 months, 2 weeks ago

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a)

\(n=a^2+4ab+b^2=(a+2b)^2-3b^2\).So \(x\) is of the form \(m^2-3n^2\). Similarly \(y\) is of the form \(p^2-3q^2\).

By Brahmgupta's identity, \(xy=(m^2-3n^2)(p^2-3q^2)=(mp+3pq)^2-3(mq+np)^2\) which is also of the form of \(n\).

Brilliant Member - 6 months, 2 weeks ago

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b) All perfect squares are of the form \(0,1 \pmod 4\). So \(x^2+4xy+y^2 \equiv 0,1,2 \pmod 4\) but \(11 \equiv 3\pmod 4\).

So no solution.

Brilliant Member - 6 months, 2 weeks ago

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