product of four consecutive odd natural numbers

[defeated_soldier]

Sixteen added to the product of four consecutive odd natural numbers is always :

(1) a prime number
(2)a perfect square
(3)a cube
(4)None of these

Ans given in my book : (2)a perfect square



MySolution :

I just took an example 1 , 3,5,7 to verify the answer. // 2 is an assumption

so, product of four consequitive odd natural numbers = 1 x 3 x 5 x 7=105

105+16=121

121 is not a perfect square.

is there any mistake in my assumption ?

 

[skeeter]

oh no? ... 121 = 11²

 

[soroban]

Hello, defeated_soldier!

Sixteen added to the product of four consecutive odd natural numbers is always :

(1) a prime number . (2) a perfect square . (3) a cube . (4) None of these

Ans given in my book: (2) a perfect square

Trying specific numbers is not a good idea
. . (expecially when you don't recognize a square when you see it).

Let \(\displaystyle x\) be the first (smallest) odd natural numbers.
Then \(\displaystyle x\,+\,2,\:x\,+\,4,\:x\,+\,6\) are the next three odd numbers.

Their product plus 16 is: \(\displaystyle \:N\;=\;x(x\,+\,2)(x\,+\,4)(x\,+\,6)\,+\,16\)

We can multiply it out and simplify,
. . and see if the result is a perfect square.


Here's a clever trick . . . not my own, of course.

We have: \(\displaystyle \:N\;=\;x(x\,+\,6)\,\cdot(x\,+\,2)(x\,+\,4)\,+\,16\)

. . \(\displaystyle N\;=\;(x^2\,+\,6x)\cdot(x^2\,+\,6x\,+\,8)\,+\,16\)

. . \(\displaystyle N\;= \;(x^2\,+\,6x\,\)+ 4 - 4\(\displaystyle )\cdot(x^2\,+\,6x\,\)+ 4 + 4\(\displaystyle )\,+\,16\)

. . \(\displaystyle N\;= \;\left[(x^2\,+\,6x\,+\,4)\,-\,4\right]\cdot\left[(x^2\,+\,6x\,+\,4)\,+\,4\right}\,+\,16\)

. . \(\displaystyle N\;=\;\left[(x^2\,+\,6x\,+\,4)^2\,-\,16\right]\,+\,16\)

. . \(\displaystyle N\;=\;(x^2\,+\,6x\,+\,4)^2\)


Therefore, \(\displaystyle N\) is always a perfect square.

 

[defeated_soldier]

i could not recognize 121 as a square number

By The way ,soroban , that was an exciting solution !

why did you took x is a odd number . how about taking 2x+1 ? of course , your solution is fine because ultimately we get the squared number form

However, but i was just thinking to assume a odd number in this form 2x+1 and others are incrementing by 2 next onwards

thanks