Is it possible that the product of two consecutive even numbers, is a perfect square? What about odd numbers?

Question

Please, don't explain it with examples.

Answer 1

It will always be one less than a perfect square.

n(n+2) = (n+1)^2 - 1

But the only time that two consecutive integers are both perfect squares is when they are 0 and 1. So if you consider 0 to be a perfect square (0*0), then 0*2 works.

Answer 2

For the product to be a suited sq., one possibility is to multiply numbers that have a similar base, yet have diverse atypical powers. the occasion you gave extra healthy here trend and a couple of is the only recommendations with that trend. 5th potential - first potential = 30 n^5 - n = 30 n^5 - n - 30 = 0 2 // . . .a million . . . 0 . . . 0 . . . 0 . . . -a million . . . -30 . . . . . . . . . . . 2 . . . 4 . . . 8 . . .sixteen . . . . 30 --------------------------------------... . . . . . . .a million . . . 2 . . .4 . . . 8 . . . 15 . . . 0 (x-2)( x^4 + 2 x^3 + 4x^2 + 8x + 15) (for the incredibly evidence you could desire to apply synthetic branch to coach possible components) yet another often is the third potential - first potential = 30 (yet differently for exhibiting evidence would be to apply incredibly opportunities) 2^3 - 2 = 6 3^3 - 3 = 24 4^3 - 4 = 60 because of the fact the version will basically get extra advantageous, you could quit; no integers extra healthy this possibility. 5th potential - third potential = 30 2^5 - 2^3 = 32 - 8 = 24 3^5 - 3^3 = 80 one+27 = fifty 4 end seventh potential - first potential = 30 2^7 - 2 = 128 - 2 end seventh potential - third potential = 30 2^7 - 2^3 = 128 - 8 = one hundred twenty end seventh potential - 5th potential = 2^7 - 2^3 = 128 - 32 = ninety six look as though this trend has not extra recommendations yet another pair is 10 and 40 which has the trend (a^3)(b) - ab = 30 perhaps a minimum of this facilitates provide you some path, yet i'm out of time for this night. If I arise with different concepts, i'm going to come decrease back.

Answer 3

Wait, I don't get why everyone is saying no. I mean, if the pair of numbers would be -2 and 0 or 0 and 2, that would be okay. Besides, it asks for product, not sum.

If we let the 2 even numbers be 2x and 2x + 2, then their product would be

2x(2x+2) = (4x^2 + 4x)

sqrt(4x^2 + 4x) = 2*sqrt(x^2+x)

Thus, it follows that it can be a perfect square if you can find other integral values for x which allows x^2+x to be square rooted. 0 would be one. The domain of this one would be x<=-1 or x>=0, thus, there may be some more.

For odd numbers, the 2 numbers would be 2x -1 and 2x +1. Thus the product is

(2x -1)(2x+1) = 4x^2-1

sqrt(4x^2-1)

There could be if you could come up with an integer x that allows 4x^2-1 to be square rooted. I think this is possible because its domain (list of acceptable values) would be x<=-1/2 or x>=1/2, though I'm not so sure.

Answer 4

No. It is not possible. suppose x is the even no or the odd number. The next even or odd no is x+2.

Their product x(x+2) can under no circumstances be a perfect square.

Answer 5

No.

x(x+2) = x^2 +2x
consider the number after this, x^2+2x+1
this is a perfect square (x+1)^2
So, x(x+2) will always be 1 short of a perfect square

Answer 6

Not possible. The root of this number (basically the geometric mean) lies between the two numbers. However the number that lies in between is odd. Its square cannot be even (product of the teo even numbers on either sides).

Answer 7

with even number it's impossible. With odd numbers a perfect example would be

1 + 3 = 4

sqrt 4 = 2

Answer 8

No this is not possible. The meaning of a square is that it is a product of LXB where L=B. Hence the answer is no.