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# If 3n + 1 is a perfect square, show that n + 1 is the sum of three perfect squares.

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If 3n + 1 is a perfect square, show that n + 1 is the sum of three perfect squares.

Nov 27, 2015

#1
+35

If 3n + 1 is a perfect square, show that n + 1 is the sum of three perfect squares.

3n + 1 is a perfect square:

$$\small{ \begin{array}{rcl} 3n+1 &=& a^2 \\ 3n &=& a^2 - 1 \\ n &=& \frac{a^2-1}{3}\\ \hline n+1 &=& \frac{a^2-1}{3} +1 \\ n+1 &=& \frac{a^2-1+3}{3}\\ n+1 &=& \frac{a^2+2}{3}\\ \end{array} }$$

Because $$\frac{a^2+2}{3}$$ is a integer then $$a^2+2$$ is divisible by 3, then $$a^2$$ is not divisible by 3  and also $$a$$ is not divisible by 3,

because if 3 is not a prime factor in $$a^2$$ a perfect spuare, then 3 is not a prime factor in  $$a$$

Two numbers are not divisible by 3. It is  $$3b +1$$ and $$3b + 2$$

1. We substitute $$a = 3b+1$$

$$\small{ \begin{array}{rcl} n+1 &=& \frac{a^2+2}{3}\qquad \text{substitute }\ a = 3b+1\\ n+1 &=& \frac{(3b+1)^2+2}{3} \\ n+1 &=& \frac{9b^2+6b+1+2}{3} \\ n+1 &=& \frac{9b^2+6b+3}{3} \\ n+1 &=& 3b^2+2b+1 \\ n+1 &=& b^2 + b^2 + b^2 +2b+1 \\ n+1 &=& b^2 + b^2 + (b+1)^2\\ \end{array} }$$

So n + 1 is the sum of three perfect squares and $$b = \frac{a-1}{3}$$.

2. We substitute $$a = 3b+2$$

$$\small{ \begin{array}{rcl} n+1 &=& \frac{a^2+2}{3}\qquad \text{substitute }\ a = 3b+2\\ n+1 &=& \frac{(3b+2)^2+2}{3} \\ n+1 &=& \frac{9b^2+12b+4+2}{3} \\ n+1 &=& \frac{9b^2+12b+6}{3} \\ n+1 &=& 3b^2+4b+2 \\ n+1 &=& b^2 + b^2 + b^2 +2b+2b+1+1 \\ n+1 &=& b^2 + b^2+2b+1 + b^2 +2b+1 \\ n+1 &=& b^2 +(b+1)^2+(b+1)^2\\ \end{array} }$$

So n + 1 is the sum of three perfect squares and $$b = \frac{a-2}{3}$$.

Example 1:

$$\small{ \begin{array}{rcl} a &=&7 \\ 3n+1 =a^2&=& 7^2 \qquad \rightarrow \qquad n=\frac{a^2-1}{3}=\frac{49-1}{3} = 16\\\\ n+1 &=& 16+1=17 \\ 17 &=& b^2+b^2+(b+1)^2 \qquad a=3b+1 \qquad \rightarrow b = \frac{a-1}{3} = \frac{7-1}{3} = 2\\ 17 &=& 2^2+2^2+3^2 = 4+4+9\\ \end{array} }$$

Example 2:

$$\small{ \begin{array}{rcl} a &=&8 \\ 3n+1 =a^2&=& 8^2 \qquad \rightarrow \qquad n=\frac{a^2-1}{3}=\frac{64-1}{3} = 21\\\\ n+1 &=& 21+1=22 \\ 22 &=& b^2+(b+1)^2+(b+1)^2 \qquad a=3b+2 \qquad \rightarrow b = \frac{a-2}{3} = \frac{8-2}{3} = 2\\ 22 &=& 2^2+3^2+3^2 = 4+9+9\\ \end{array} }$$ Nov 27, 2015
edited by heureka  Nov 27, 2015
edited by heureka  Nov 27, 2015

#1
+35

If 3n + 1 is a perfect square, show that n + 1 is the sum of three perfect squares.

3n + 1 is a perfect square:

$$\small{ \begin{array}{rcl} 3n+1 &=& a^2 \\ 3n &=& a^2 - 1 \\ n &=& \frac{a^2-1}{3}\\ \hline n+1 &=& \frac{a^2-1}{3} +1 \\ n+1 &=& \frac{a^2-1+3}{3}\\ n+1 &=& \frac{a^2+2}{3}\\ \end{array} }$$

Because $$\frac{a^2+2}{3}$$ is a integer then $$a^2+2$$ is divisible by 3, then $$a^2$$ is not divisible by 3  and also $$a$$ is not divisible by 3,

because if 3 is not a prime factor in $$a^2$$ a perfect spuare, then 3 is not a prime factor in  $$a$$

Two numbers are not divisible by 3. It is  $$3b +1$$ and $$3b + 2$$

1. We substitute $$a = 3b+1$$

$$\small{ \begin{array}{rcl} n+1 &=& \frac{a^2+2}{3}\qquad \text{substitute }\ a = 3b+1\\ n+1 &=& \frac{(3b+1)^2+2}{3} \\ n+1 &=& \frac{9b^2+6b+1+2}{3} \\ n+1 &=& \frac{9b^2+6b+3}{3} \\ n+1 &=& 3b^2+2b+1 \\ n+1 &=& b^2 + b^2 + b^2 +2b+1 \\ n+1 &=& b^2 + b^2 + (b+1)^2\\ \end{array} }$$

So n + 1 is the sum of three perfect squares and $$b = \frac{a-1}{3}$$.

2. We substitute $$a = 3b+2$$

$$\small{ \begin{array}{rcl} n+1 &=& \frac{a^2+2}{3}\qquad \text{substitute }\ a = 3b+2\\ n+1 &=& \frac{(3b+2)^2+2}{3} \\ n+1 &=& \frac{9b^2+12b+4+2}{3} \\ n+1 &=& \frac{9b^2+12b+6}{3} \\ n+1 &=& 3b^2+4b+2 \\ n+1 &=& b^2 + b^2 + b^2 +2b+2b+1+1 \\ n+1 &=& b^2 + b^2+2b+1 + b^2 +2b+1 \\ n+1 &=& b^2 +(b+1)^2+(b+1)^2\\ \end{array} }$$

So n + 1 is the sum of three perfect squares and $$b = \frac{a-2}{3}$$.

Example 1:

$$\small{ \begin{array}{rcl} a &=&7 \\ 3n+1 =a^2&=& 7^2 \qquad \rightarrow \qquad n=\frac{a^2-1}{3}=\frac{49-1}{3} = 16\\\\ n+1 &=& 16+1=17 \\ 17 &=& b^2+b^2+(b+1)^2 \qquad a=3b+1 \qquad \rightarrow b = \frac{a-1}{3} = \frac{7-1}{3} = 2\\ 17 &=& 2^2+2^2+3^2 = 4+4+9\\ \end{array} }$$

Example 2:

$$\small{ \begin{array}{rcl} a &=&8 \\ 3n+1 =a^2&=& 8^2 \qquad \rightarrow \qquad n=\frac{a^2-1}{3}=\frac{64-1}{3} = 21\\\\ n+1 &=& 21+1=22 \\ 22 &=& b^2+(b+1)^2+(b+1)^2 \qquad a=3b+2 \qquad \rightarrow b = \frac{a-2}{3} = \frac{8-2}{3} = 2\\ 22 &=& 2^2+3^2+3^2 = 4+9+9\\ \end{array} }$$ heureka Nov 27, 2015
edited by heureka  Nov 27, 2015
edited by heureka  Nov 27, 2015
#2
+5

Very nice, heureka.......!!!!   Nov 27, 2015