Number Theory
# Quadratic Diophantine Equations

\(4\) has the property that if one adds it to double its square, it yields a perfect square. In other words for natural numbers \(m,n\):

\[n^2 + n^2 + n = m^2 \]

There are four such \(n<10^6 \). If \(4\) is the smallest \(n\), what is the second smallest \(n\)?

What is the largest integer \(\displaystyle n\) for which \( n^2+24n+16\) is a perfect square?

\(365\) can be written as a sum of \(2\) consecutive perfect squares and also \(3\) consecutive non-zero perfect squares: \[365=14^2+13^2=12^2+11^2+10^2\]

What is the next number with this property?

For how many positive integers \(n<10^6\) is \(2\times n! \times (n+2)!\) a perfect square?

Find the sum of all positive integers \(m\) such that \(2^m\) can be expressed as sums of four factorials (of positive integers).

**Details and assumptions**

The number \( n!\), read as **n factorial**, is equal to the product of all positive integers less than or equal to \(n\). For example, \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\).

The factorials do not have to be distinct. For example, \(2^4=16\) counts, because it equals \(3!+3!+2!+2!\).

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