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# What square is the product of four consecutive odd integers ?

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What square is the product of four consecutive odd integers ?

Dec 8, 2015

#1
+23299
+15

What square is the product of four consecutive odd integers ?

$$\begin{array}{rcl} (2n-3)(2n-1)(2n+1)(2n+3) &=& x^2 \\ (2n-3)(2n+3)(2n-1)(2n+1) &=& x^2 \\ (4n^2-3^2)(4n^2-1) &=& x^2 \\ 16n^4-40n^2+3^2 &=& x^2 \qquad n = 0\\ 3^2 &=& x^2 \\ x &=& 3 \\\\ (2\cdot 0-3)(2\cdot 0-1)(2\cdot 0+1)(2\cdot 0+3) &=& 3^2 \\ (-3)\cdot (-1)\cdot (1)\cdot (3) &=& 3^2 \\ \end{array}$$

Dec 8, 2015
edited by heureka  Dec 8, 2015
edited by heureka  Dec 8, 2015

#1
+23299
+15

What square is the product of four consecutive odd integers ?

$$\begin{array}{rcl} (2n-3)(2n-1)(2n+1)(2n+3) &=& x^2 \\ (2n-3)(2n+3)(2n-1)(2n+1) &=& x^2 \\ (4n^2-3^2)(4n^2-1) &=& x^2 \\ 16n^4-40n^2+3^2 &=& x^2 \qquad n = 0\\ 3^2 &=& x^2 \\ x &=& 3 \\\\ (2\cdot 0-3)(2\cdot 0-1)(2\cdot 0+1)(2\cdot 0+3) &=& 3^2 \\ (-3)\cdot (-1)\cdot (1)\cdot (3) &=& 3^2 \\ \end{array}$$

heureka Dec 8, 2015
edited by heureka  Dec 8, 2015
edited by heureka  Dec 8, 2015
#2
+105606
+10

What square is the product of four consecutive odd integers ?

$$(2n-3)(2n-1)(2n+1)(2n+3)=k^2\\ (4n^2-9)(4n^2-1)=k^2\\ 16n^4-40n^2+9=k^2\\ 16n^4-40n^2=k^2-9\\ 16(n^4-\frac{40}{16}n^2)=k^2-9\\ 16(n^4-\frac{10}{4}n^2+\frac{25}{16})=k^2-9+25\\ 16(n^2-\frac{5}{4})^2=k^2+16\\ 16(n^2-\frac{5}{4})^2=k^2+16\\$$

We have a pythagorean triad here.

I mean      k,      4     and    4(n^2-5/4)      is a pythagorean triad

If k=3 then

$$16(n^2-\frac{5}{4})^2=25\\ 4(n^2-\frac{5}{4})=5\\ (n^2-\frac{5}{4})=\frac{5}{4}\\ (n^2-\frac{5}{4})=\frac{5}{4}\\ n^2=0\\ n=0$$

So one solution is      $$-3*-1*1*3=9$$

YES I know, I did that the SUPER long way!

According to WolframAlpha, this is the ONLY integer solution.

So 9 is the product of 4 consecutive odd integers.

Dec 8, 2015