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

math grad - Interest: Number theory
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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.