## a = √(√(n) + √((k*n)+1))

Can you find integer solutions of

This can be viewed as a Pell Equation

$a^2 \; = \; \sqrt {n} \; + \; \sqrt {(k \, n) + 1}$

in integers, so we know that there exist integers   $x$   and   $y$   such that:

$n \; = \; y^2$
$k \, n \; + \; 1 \; = \; x^2$
$x \; + \; y \; = \; a^2$

$k \, y^2 \; + \; 1 \; = \; x^2$

math grad - Interest: Number theory
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### 2 Responses to a = √(√(n) + √((k*n)+1))

1. paul says:

There are an infinite number of solutions to that. There are infinite sets when n=1 and n=4 and n=9, it seems to miss 16, back at 25 and 36. There is an interesting relationship at n=36, the value of “a” is defined by the pairs of 18t+-5, which goes {{5}, {13, 23}, {31,41},{49,59},{67,77},{85,95},…}
Here are some values at n=36, format is {n, k, a}

{36,10,5}
{36,738,13}
{36,7598,23}
{36,25334,31}
{36,77934,41}
{36,159334,49}
{36,335434,59}
{36,558258,67}
{36,974498,77}
{36,1447610,85}

{36,288372401010,1795}..t=100
{36,294852439010,1805}..t=100

The last two were found by solving back for k

Paul.