Let x be first number and y- second number. When you subtract first square number from second you get \(y^2-x^2\) and it is equal to 7.
Hence
\(y^2-x^2=7\).
Find factors of right and left sides:\((y-x)(y+x)=1\cdot 7\).
Since \(y+x\ \textgreater \ y-x\) you have that
\(\left \{ {{y-x=1} \atop {y+x=7}} \right.\).
Add these equations:
\(y-x+y+x=1+7, \\ 2y=8, \\ y=4\).
Then x=7-y=7-4=3.
Answer: \(4^2-3^2=7.\)
