a consecutive number that when multiplied together equals a square number?

Answer 1

WTF? - Your question makes no sense. I suppose you mean TWO consecutive numbers multiplied together...

But in that case the answer is impossible:

If the two consecutive numbers are A and A+1

A x A+1 will be greater than A^2 and less than (A+1)^2 i.e. it will always be between two squares.

:-/

.

Answer 2

1 and 2

Answer 3

There is no number 'X' where X * (X+1) generates a number that is an exact square. The reason is that the function sqrt(X*(X+1)) tends to X + 0.5 as X approaches infinity. The same is true for sqrt(X * (X-1)), this approaches |X|+0.5 as X approaches -infinity. The only possible answer is 0 and 1 which produces 0 as the answer, but I'm not too sure if 0 is a true square.

Interestingly, sqrt(X * (X+2)) tends to X+1 as X approaches infinity, so there is an answer here of infinity and infinity+1 !

The general expression here then is sqrt(X * (X+y)) tends to X+0.5 if y is odd, and X+1 if y is even.

Answer 4

consecutive numbers 3 and 4
3^2 = 9
4^2 = 16

added together = 25 = 5^2

Answer 5

You must mean two consecutive numbers. Can only be 1 and 0

Answer 6

Nothing unless you consider 0 a square number. If so, 0 and 1

Answer 7

I'm not sure what you mean by *a* consecutive number; that doesn't make sense. Perhaps you mean two consecutive integers; in which case 0 and 1 would work.

Answer 8

-1 *0 =0
or
0*1 =0

Other than that, there are none.
If you have x and (x+1)
the squareroot of x*(x+1) will be between x and x+1

Answer 9

I think you mean

x(x+1) = y^2
or x^2+x = y^2


The only solution That I think of is
x=0 or -1

Answer 10

If you're talking about integers, there isn't 1.