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How do I prove that product of $4$ consecutive integers is equal to 1 less than squared of the average of the product of last and middle terms ?

Ex: let $a,b,c,d$ be $4$ consecutive integers, then: $$abcd= \left( \dfrac{ad+bc}{2} \right)^{2}-1$$

I can do this just by supposing values of $a= some\ even /odd$ but I need to arrive at this by not in this way, please help.

I found something similar here but how to frame it in this way?

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mnulb
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  • The question you link to just shows that $abcd$ is one less than a square. You can follow the algebra to find what it is the square of, which shows what you are after. – Ross Millikan Dec 11 '16 at 16:03
  • Sorry actually I m new here so I have little problem in formatting. But I will improve it. – Kanwaljit Singh Dec 11 '16 at 16:04
  • @RossMillikan that's what I can't do, need MSE help. – mnulb Dec 11 '16 at 16:31
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    The critical idea is using the fact that $a,b,c,d$ are consecutive, which is where replacing them with $x+1,x+2,x+3,x+4$ comes from. Having done that, Kanwaljit Singh's solution is expanding the polynomials. Look at each step carefully and make sure you understand that it works. – Ross Millikan Dec 11 '16 at 16:43
  • I tried to frame that with more explanation. Hope it helps. – Kanwaljit Singh Dec 11 '16 at 16:44

1 Answers1

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Let numbers are x, x+1, x+2, x+3.

Then Left side,

x(x+1)(x+2)(x+3)

$=x(x+3)(x+1)(x+2)$

$= (x^2+3x)(x^2+3x+2)$

Put x^2+3 = Z.

$= Z(Z+2) = Z^2+2$

Now right side

$= \left(\frac{(x)(x+3)+(x+1)(x+3)}{2}\right)^{2} - 1$

$= \left(\frac{(x^2+3x)+(x^2+3x+2)}{2}\right)^{2} - 1$

$= \left(\frac{(Z)+(Z+2)}{2}\right)^{2} - 1$

$= \left(\frac{2Z+2}{2}\right)^{2} - 1$

$= (Z+1)^2 - 1$

$= Z^2+2Z+1-1$

$= Z^2+2Z$

Kanwaljit Singh
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