If 3n + 1 is a perfect square, show that n + 1 is the sum of three perfect squares.
+26367 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:
\(\small{ \begin{array}{rcl} a &=&8 \\ 3n+1 =a^2&=& 8^2 \qquad \rightarrow \qquad n=\frac{a^2-1}{3}=\frac{64-1}{3} = 21\\\\ n+1 &=& 21+1=22 \\ 22 &=& b^2+(b+1)^2+(b+1)^2 \qquad a=3b+2 \qquad \rightarrow b = \frac{a-2}{3} = \frac{8-2}{3} = 2\\ 22 &=& 2^2+3^2+3^2 = 4+9+9\\ \end{array} }\)
+26367 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:
\(\small{ \begin{array}{rcl} a &=&8 \\ 3n+1 =a^2&=& 8^2 \qquad \rightarrow \qquad n=\frac{a^2-1}{3}=\frac{64-1}{3} = 21\\\\ n+1 &=& 21+1=22 \\ 22 &=& b^2+(b+1)^2+(b+1)^2 \qquad a=3b+2 \qquad \rightarrow b = \frac{a-2}{3} = \frac{8-2}{3} = 2\\ 22 &=& 2^2+3^2+3^2 = 4+9+9\\ \end{array} }\)
