NT-B1-M05-P023 Difference of Squares with a Shift

After factoring the left-hand side, factor out \(x+y\).

Factor: \((x-y)(x+y)=2(x+y)+15\). Hence \((x+y)(x-y-2)=15\). Let \(A=x+y\), \(B=x-y-2\). Then \(AB=15\), \(A>0\), \(B>0\). Check the pairs \((15,1)\), \((5,3)\), \((3,5)\), \((1,15)\). Since \(y=\frac{A-B-2}{2}\), a positive \(y\) occurs only for \((A,B)=(15,1)\). Then \(y=6\), \(x=9\). The answer is \((9,6)\).

Here factorisation requires grouping the right-hand side first.