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.

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TopNewestc) \(x^2+4xy+y^2=(x+2y)^2-3y^2\) 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. – Svatejas Shivakumar · 4 months, 1 week 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 \(n\). – Svatejas Shivakumar · 4 months, 1 week 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. – Svatejas Shivakumar · 4 months, 1 week ago

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