How to find integer solutions an equation under a squareroot

Question

I have a function $f(z) = \sqrt{(2z^2+1)^2+ 2^2 z(102z^2+151z+51)} $

I know it has some solutions where if $z$ and $f(z)$ is an integer, but are there infinitely many? or maybe is there some $z = f(c)$ such that $f(z)$ is always an integer?

My initially try was to try and set the function under the radical equal to a Pythagorean triple, but Im struggleing.

So yeh any ideas? (also I know there are solutions for $z = 0, 1, 12, 62412,$ and maybe $124887$ (I think my Matlab program was rounding incorrectly for this last one) but then I couldn't find any more solutions

Answer 1

Your $z = 62412$ is incorrect.

62412  60791323154235277969
7796879065
60791323154235274225
60791323154235277969
3744

Taking your $$ g(z) = 4z^4 + 408z^3 + 608z^2 + 204z + 1, $$ we get $$ g(z) = ( 2 z^2 + 102 z - 2449 )^2 + 499800z - 5997600 = ( 2 z^2 + 102 z - 2449 )^2 + 499800(z -12) . $$ Also $$ g(z) = ( 2 z^2 + 102 z - 2448 )^2 - (4z^2 - 499596z + 5992703) $$

As a result, for $$ z \geq 124888, $$ we have $$ ( 2 z^2 + 102 z - 2449 )^2 < g(z) < ( 2 z^2 + 102 z - 2448 )^2 $$ so $g(z)$ cannot be a square for integer $z \geq 124888$

parisize = 4000000, primelimit = 500509
? g = ( 2 * z^2 + 1)^2 + 4 * z * (102 * z^2 + 151 * z + 51)
%1 = 4*z^4 + 408*z^3 + 608*z^2 + 204*z + 1

? ( 2 * z^2 + 102 * z )^2
%2 = 4*z^4 + 408*z^3 + 10404*z^2

? ( 2 * z^2 + 102 * z - 2450 )^2
%13 = 4*z^4 + 408*z^3 + 604*z^2 - 499800*z + 6002500

? ( 2 * z^2 + 102 * z - 2449 )^2
%14 = 4*z^4 + 408*z^3 + 608*z^2 - 499596*z + 5997601


? g - ( 2 * z^2 + 102 * z - 2449 )^2
%15 = 499800*z - 5997600


? polroots( 499800*z - 5997600 )
%18 = [12.00000000000000000000000000 + 0.E-28*I]~
? 



? g - ( 2 * z^2 + 102 * z - 2448 )^2
%16 = -4*z^2 + 499596*z - 5992703


? polroots(  -4*z^2 + 499596*z - 5992703  )
%17 = [11.99625025036284782655713999 + 0.E-28*I, 124887.0037497496371521734429 + 0.E-28*I]~
?