Four consecutive terms of an AP

 
We know that if you multiply any four consecutive positive integers and add 1 to the product, you’ll get a square number.

(x - 1) \, x \, (x + 1) \, (x + 2) \; + \; 1
= \; x^4 \; + \; 2 \, x^3 \; - x^2 \; - \; 2 \, x \; + \; 1
= \; (x^2+x-1)^2
 

The product of four consecutive terms of an arithmetic progression added to the fourth power of the common difference is always a perfect square

(x - n) \, x \, (x + n) \, (x + 2n) \; + \; n^4
= \; n^4 \; - \; 2 \, n^3 \, x \; - \; n^2 \, x^2 \; + \; 2 \, n \, x^3 \; + \; x^4
= \; (n^2 - n \, x - x^2)^2

 

This also applies to Rational numbers in general. Find few examples

 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

One Response to Four consecutive terms of an AP

  1. Paul says:

    Staring with x = 1/2 to x = 9/10 in steps of 1/10 and n = 1/2 to n = 7/8 in steps of 1/8.
    Format is {square No, Root of No}, {x, n}

    {1/16,1/4} , {1/2,1/2}
    {121/4096,11/64} , {1/2,5/8}
    {1/256,1/16} , {1/2,3/4}
    {25/4096,5/64} , {1/2,7/8}
    {1681/10000,41/100} , {3/5,1/2}
    {303601/2560000,551/1600} , {3/5,5/8}
    {9801/160000,99/400} , {3/5,3/4}
    {36481/2560000,191/1600} , {3/5,7/8}
    {3481/10000,59/100} , {7/10,1/2}
    {737881/2560000,859/1600} , {7/10,5/8}
    {32761/160000,181/400} , {7/10,3/4}
    {290521/2560000,539/1600} , {7/10,7/8}
    {6241/10000,79/100} , {4/5,1/2}
    {1437601/2560000,1199/1600} , {4/5,5/8}
    {73441/160000,271/400} , {4/5,3/4}
    {844561/2560000,919/1600} , {4/5,7/8}
    {10201/10000,101/100} , {9/10,1/2}
    {2468041/2560000,1571/1600} , {9/10,5/8}
    {136161/160000,369/400} , {9/10,3/4}
    {1771561/2560000,1331/1600} , {9/10,7/8}

    Paul.

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