If n is a positive integer such that 2n + 1 is a perfect square, show that n + 1 is the sum of two successive perfect squares.

If n is a positive integer such that 2n + 1 is a perfect square, show that n + 1 is the sum of two successive perfect squares.

2n + 1 is a perfect square:

\(\begin{array}{rcl} 2n+1 &=& a^2 \\ 2n &=& a^2 - 1 \\ n &=& \frac{a^2-1}{2}\\ \hline n+1 &=& \frac{a^2-1}{2} +1 \\ n+1 &=& \frac{a^2-1+2}{2}\\ n+1 &=& \frac{a^2+1}{2}\\ \end{array}\)

Because \(\frac{a^2+1}{2}\) is a integer then \(a^2+1\) is even and divisible by 2, then \(a^2\) is odd or also \(a\) is odd.

A odd number is  \(2b+1\)

\(\begin{array}{rcl} n+1 &=& \frac{a^2+1}{2}\qquad \text{substitute }\ a = 2b+1\\ n+1 &=& \frac{(2b+1)^2+1}{2} \\ n+1 &=& \frac{4b^2+4b+1+1}{2} \\ n+1 &=& \frac{4b^2+4b+2}{2} \\ n+1 &=& 2b^2+2b+1 \\ n+1 &=& b^2 + b^2+2b+1 \\ n+1 &=& b^2 +(b+1)^2\\ \end{array}\)

So n + 1 is the sum of two successive perfect squares and \(b = \frac{a-1}{2}\).

Example:

\(\begin{array}{rcl} a &=& 5\\ 2n+1 =a^2 &=& 25 \qquad \rightarrow\qquad n = \frac{a^2-1}{2} = \frac{25-1}{2} = 12\\\\ b &= &\frac{a-1}{2} = \frac{5-1}{2} = 2\\ n+1 &=& 12+1 = 13 \\ 13 &=& b^2 + (b+1)^2 = 2^2+3^2 = 4+9\\ \end{array}\)

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