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$n^{2}+20n+15=n^{2}+2.10.n+100-85=(n+10)^{2}-85$

Let,

$(n+10)^{2}-85=m^{2}$

or,$(n+10)^{2}-m^{2}=85$

or, $ (n+10+m)(n+10-m)=85$

we can write ,

$85=17*5 ; 85=1*85 ;85=(-1)*(-85); 85=(-17)*(-5)$

So, let try with first one $85=17*5$ ,

$n+10+m=17$

$n+10-m=5$

So, $2n+20=22$

or,  $n=1$

So , let see $n=1$ gives a perfect square or not?

$1^{2}+1*20+15=36=6^{2}$

Now let try with , $85=(-17)(-5)$

$n+10+m=-17$

$n+10-m=-5$

so, $2n+20=-22$

so, $n=-21$

So , let see $n=-21$ gives a perfect square or not?

=>$(-21)^{2}+(-21)*20+15=36=6^{2}$

Now let try with , $85=(85)(1)$

$n+10+m=85$

$n+10-m=1$

so, $2n+20=86$

so, $n=33$

So , let see $n=33$ gives a perfect square or not?

=>$(33)^{2}+(33)*20+15=1764=42^{2}$

Now let try with , $85=(-85)(-1)$

$n+10+m=-85$

$n+10-m=-1$

so, $2n+20=-86$

so, $n=-53$

So , let see $n=33$ gives a perfect square or not?

=>$(-53)^{2}+(-53)*20+15=1764=42^{2}$

So, The Here 4 solutions are possibles which are $1,-21,33,-53$ .

Check this comment here for some more insight .

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