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 Three in a row (Posted on 2016-02-08)
Prove that there exist infinitely many integers n such that n, n+1, n+2 are each a sum of two squares of integers.

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 easy with 0 Comment 3 of 3 |
Admitting the digit 0 as a square the problem is quite easy.

Consider a square n=m^2

Substracting 2m-1 and adding an square a^2 (just taking care that a^2 is even and its value is one less than 2m-1), the problem is done.

2m-1=a^2+1, so m=(a^2+2)/2 and we have the second number, composed by the squares of m-1 and a. This is true for each value of a (a is even).

The third number is always n+1.

Ex (to begin assign a value to "a"):
a=14 m=99 n=9801
98^2+14^2=9800
99^2+1^2=9802

But if 0 is not admitted as a square I don't know if the ennunciate is true.

Edited on February 11, 2016, 4:32 am
 Posted by armando on 2016-02-11 04:15:29

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