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Verified Answer :

LammettHash

\(n^2+(n+1)^2=2n^2+2n+1=221\)
\(2n^2+2n=220\)
\(n^2+n=110\)
\(n^2+n-110=0\)
\((n-10)(n+11)=0\implies n=10,n=-11\)

But since \(n\) must be a whole number, we ignore \(n=-11\). So the two integers are \(n=10\) and \(n+1=11\).

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