SOLUTION: Find three consecutive even integers such that the sum of the squares of the first and second integers is equal to the square of the third integer.

Question 245451: Find three consecutive even integers such that the sum of the squares of the first and second integers is equal to the square of the third integer.
Found 2 solutions by richwmiller, solver91311:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
let n= the first integer n+1= the second integer and n+2= the third integer
n^2+(n+1)^2=(n+2)^

work out the squares
(n+1)^2=n^2+2n+1
(n+2)^2=n^2+4n+4
n^2+ n^2+2n+1=n^2+4n+4
subtract n^2 from both sides
n^2+2n+1=+4n+4
subtract 4n from both sides
n^2-2n+1=4
subtract 4 from both sides
n^2-2n-3=0
n=3 and -1
try 3, 4 and 5
3^2+4^2=5^2
9+16=25
ok
try -1,0,1
(-1)^2+0^2=1^2
1=1
ok

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Let represent the first integer.

Then is the next consecutive integer. But if is even, then , must be odd. Therefore, the next consecutive even integer must be .

It follows then that the next consecutive integer after that must be

Square the first integer:

Square the second integer:

Add these two quantities:

And this sum is equal to the square of the third integer:

Collect like terms in the LHS, leaving the RHS equal to zero:

Now solve the factorable quadratic. Each of the roots will be the first of a series of three consecutive even integers that fit the parameters of the problem. Remember that 0 is a perfectly good even integer in this context.

John

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