Finding $n$: $\sqrt {n+16}\,\, and\,\, \sqrt {16n+1} \in N$

[Albert1]

$n\in N$
$\sqrt {n+16}\,\, and\,\, \sqrt {16n+1}$ are also $\in N$
find $n$

 

[kaliprasad]

$n\in N$
$\sqrt {n+16}\,\, and\,\, \sqrt {16n+1}$ are also $\in N$
find $n$

Spoiler

let $n-=m^2-16$
then $16n + 1 = k^2$ as it is a perfect square
or $16m^2-255 = k^2$
or $16m^2-k^2 = 255$
or $(4m-k)(4m+k)= 255 = 1 * 255 = 3 * 85 = 5 * 51 = 15 * 17$
choosing pairs
$4m-k = 1, 4m+ k = 255$ we get m = 32 or n = 1008
$4m-k = 3, 4m+ k = 85$ we get m = 11 or n = 105
$4m-k = 5, 4m+ k = 51$ we get m = 7 or n = 33
$4m-k = 15, 4m+ k = 17$ we get m = 8 or n = 0 which is not admissible as n is natual number

so n = 33 or 105 or 1008

 

[Albert1]

Spoiler

let $n-=m^2-16$
then $16n + 1 = k^2$ as it is a perfect square
or $16m^2-255 = k^2$
or $16m^2-k^2 = 255$
or $(4m-k)(4m+k)= 255 = 1 * 255 = 3 * 85 = 5 * 51 = 15 * 17$
choosing pairs
$4m-k = 1, 4m+ k = 255$ we get m = 32 or n = 1008
$4m-k = 3, 4m+ k = 85$ we get m = 11 or n = 105
$4m-k = 5, 4m+ k = 51$ we get m = 7 or n = 33
$4m-k = 15, 4m+ k = 17$ we get m = 8 or n = 0 which is not admissible as n is natual number

so n = 33 or 105 or 10008

so n = 33 or 105 or 10008 (a typo)
n=33,105,1008