we can write $n^2-n-2=(n-2)(n+1)$ which should be divisible by 16 and $n^2+2n-3=(n+3)(n-1)$ which should be divisible by 25
for divisible by 16 ,number should be 112,128,144,160,176,192,208,224...288.for 100<n<300.
now we to choose such no. between 100 and 300 such that any factor of these two $ (n-2)(n+1)$ will became factor of 16 which will greater than 100.
for example if we want to make any of these two $ (n-2)(n+1)$ as 112 then n should be either 114 or 111 and these no.'s also be satisfy divisible by 25 then only we consider that number so checking n=114 in $( n+3)(n-1)$ =117*113 which is not divisible by 25 same as can't take 111 also.
next ,to make any of these two $ (n-2)(n+1)$ as 128 then n should be either 130 or 131 .putting these numbers in $( n+3)(n-1)$=133*129 or 134*130 both are not divisible by 25.
like this we keep try on next 144 ....224..
only one no. is possible if we want to make any of these two $ (n-2)(n+1)$ as 224 then should be either 226 or 223 ,now these no.'s also be satisfy divisible by 25 then only we consider that number so checking n=226 in $( n+3)(n-1)$=229*225 which is divisible by 25 and n=223 will not satisfy .
so only n=226 is possible.
