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# 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.

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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.

Guest Nov 26, 2015

#1
+18715
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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:

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

heureka  Nov 26, 2015
Sort:

#1
+18715
+15

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:

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

heureka  Nov 26, 2015
#2
+78755
+5

Vey nice, heureka......!!!!!

CPhill  Nov 26, 2015

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