Three consecutive odd integers are such that the square of the third integer is 33 less than the sum of the squares of the first two. One solution is −5, −3,
Mathematics
colemason4898
Question
Three consecutive odd integers are such that the square of the third integer is 33 less than the sum of the squares of the first two. One solution is −5, −3, and−1. Find three other consecutive odd integers that also satisfy the given conditions.
1 Answer

1. User Answers LammettHash
If [tex]x[/tex] is the first integer, then
[tex](x+4)^2=x^2+(x+2)^233\iff x^24x45=(x+5)(x9)=0[/tex]
The other possibility is then [tex]x=9[/tex], so the other two integers would be [tex]x+2=11[/tex] and [tex]x+4=13[/tex].