We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#1**+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

#1**+15 **

Best Answer

**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

#2**+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.

Melody Dec 8, 2015