Find three consecutive integers such that the first times the third is 1 less than the second squared.?

Answer 1

1 2 3
2 3 4
3 4 5
4 5 6
wait, there's a pattern here. does it work for all numbers?

x, x+1, x+2

(x+1)^2 (the middle number squared)
= x^2 + 2x + 1
product of first and third numbers =
x(x+2) = x^2 + 2x
which is always 1 less than
x^2 + 2x + 1.

does it also work for negative numbers?
let's try -5, -4, -3
-5 * -3 = 15
(-4)^2 = 16
seems to work. let's prove it.
for x > 0, let's examine -x-1, -x, -x+1.
(-x)^2 = x^2
(-x-1)(-x+1) = x^2 -x +x -1 = x^2 - 1,
one less than x^2, quod erat dictum.
how about if we cross zero?
-1, 0, 1
(-1)*1 = -1
0^2 = 0
yes, that works too!

Answer 2

This is actually true for *any* integers (even negative ones).

For example:
1, 2, 3:
1 * 3 = 3
2² - 1 = 4 - 1 = 3

99, 100, 101:
99 * 101 = 9999
100² - 1 = 10000 - 1 = 9999

-1, 0, 1:
-1 * 1 = -1
0 - 1 = -1

-5, -6, -7:
-5 * -7 = 35
(-6)² - 1 = 36 - 1 = 35

The reason is pretty obvious if you try to figure this out using algebra.

Let the numbers be:
x = first integer
x + 1 = second integer
x + 2 = third integer

So you are trying to find solutions for:
x (x + 2) = (x + 1)² - 1

Expanding out the left and right sides:
x² + 2x = x² + 2x + 1 - 1

Simplifying:
x² + 2x = x² + 2x

This is true for all values of x. So *any* set of consecutive integers will work.

The answer is as easy as 1-2-3!

Answer 3

Suppose the three numbers are
n-1, n and n+1
(n-1)(n+1) = n^2 -1
This is true for any n.
So
three consecutive integers such that the first times the third is 1 less than
the second squared
is always true.

Th

Answer 4

This works with any three consecutive non-negative integers. Just choose any three numbers that are consecutive and at least zero.

4, 5, 6
4*6 = 24
5^2 = 25

11, 12, 13
11*13 = 143
12^2 = 144

Why does this work?

Let the three consecutive numbers be n-1, n, and n+1

The product of the first and the third is (n-1)(n+1).
If you multiply this out you get n^2 - 1, which is one less than n^2.

Answer 5

What about 1,2,3? The first times the third 1*3=3 which is one less than the second squared 2^2=4-1=3.

Answer 6

ANY three consecutive integers will work!! (Everyone until now has specified positive or non-negative. They work too.
{-3,-4,-5} : (-3)(-5) = (-4)^2 - 1; 15=15. In fact, it also works for any three REAL numbers that are 1 apart {π, π+1, π+2}. There is nothing in the proof that precludes negativity.)

Let them be x, x+1, and x+2
x(x+2) always equals (x+1)^2 - 1

Proof:
x(x+2) = x^2 + 2x
= x^2 + 2x + 1 - 1
= (x+1)^2 - 1

{1,2,3}
1*3 = 2^2 -1

{2,3,4}
2*4 = 3^2 -1

etc.!!!

Answer 7

The integers are x, x+1 and x+2.

So:
(x)(x+2) < (x+1)²
x² + 2x < x² + 2x + 1

Choose any positive integer and it will work.

ie. x = 1
1, 2, 3

1² + 2(2) = 5
1² + 2(2) + 1 = 6

And 5 is less than 6.

Answer 8

2 - 3 - 4.
2*4 EQUAL TO (3*3)-1 = 8.