Doesn't that make them all true?

If $n = -1$ , then $n \times n = 1$ , $n \times n \times n = -1$ , and $n \times n \times n \times n = 1$ ; which means $\boxed{n \times n \times n = 1}$ is false.

For completeness, we can show that $n \times n \times n = 1$ is the only equation that can be false with the other equations being true:

The equation $n \times n = 1$ cannot be the only false equation because then $n \times n \times n = 1$ and $n \times n \times n \times n = 1$ , which by substitution means $n \times (n \times n \times n) = n \times 1 = 1$ or $n = 1$ , which would make $n \times n = 1$ a true equation, which contradicts our assumption.

Likewise, the equation $n \times n \times n \times n = 1$ cannot be the only false equation because then $n \times n = 1$ , and $n \times n \times n \times n = (n \times n) \times (n \times n) = 1 \times 1 = 1$ , which would make $n \times n \times n \times n = 1$ a true equation, which also contradicts our assumption.

Therefore, $n \times n \times n = 1$ is the only equation that can be false with the other equations being true.

If n=1, all are true. If n=-1, then you need to use the rule for multiplying negatives: two negatives=positive, and negative x positive = negative. The result, when n=-1, is that n x n x n= -n

We will assume that the question has a consistent answer and continue with a proof by contradiction.

* Suppose that the middle equation, $n^3=1$ is true. *

This has only one real solution, $n=1$

But $n=1$ makes the other two equations true as well!

This contradicts the condition that precisely one equation is false.

So we must negate the starred supposition and say that the equation $n^3=1$ must be false.

$n*n=1\quad \Longleftrightarrow \quad n=1\quad or\quad n=-1\\ n*n*n*n={ n }^{ 4 }=1\quad \Longleftrightarrow \quad { n }^{ 2 }=1\quad or\quad { n }^{ 2 }=-1\\ \qquad \qquad \qquad \qquad \Longleftrightarrow \quad n=-1\quad or\quad n=1\quad \\ \qquad \qquad \qquad \qquad \Longleftrightarrow \quad n*n=1\\ Then\quad the\quad first\quad and\quad the\quad last\quad equations\quad are\quad the\quad same.\quad So\quad the\quad second\quad is\quad the\quad only\quad possible\quad false\quad equation.\\ (For\quad verification\quad n*n*n=1\Longleftrightarrow { n }^{ 3 }=1\quad \Longleftrightarrow \quad n=1\quad \\ Then\quad the\quad (-1)\quad is\quad the\quad counter\quad example\quad for\quad the\quad second)$

n = -1 simple as that

This problem is indeed beautiful. As we know if A=x+iy and if y=0 we call it real number. Let's use this we see n^2-1=0 , n^3-1=0 and n^4-1=0 Form this (https://drive.google.com/file/d/1kTZK6-0_Pufeel4OGFwYVoyBdI5YcfMi/view?usp=drivesdk) Read the PDF first, then we see there are 2 common roots there between n^2 and n^4 but they are not there for n^3 so now we can see easily that for n=1 and n=-1 we will get 1st and 3rd statement true but not the 2nd one.

I was thinking, 'How could a number times multiple times become 1?' So, since it's a real number it has to be based on 1. But, that would satisfy all equations, so -1 is the other option.

Three equations in general was n^2=1, n^3=1 n^4=1

The first and the last one brings us all the real solutions. But the second one could bring complex one! And always a cubic rest cant be positive!

That's why it may be false!

Before it dawned on me what n is, I realized that, if n x n = 1, then n x n x n x n is equivalent to 1 x 1, which is of course right. Assuming the first equation to be true meant the third equation is also true, leaving the second as the one that’s false.

Let n=-1 (-1)(-1)=1 1=1 which is a true statement therefore (n)(n) is also true (-1)(-1)(-1)(-1)=1 (1)(1)=1 1=1 which is also true therefore the statement (n)(n)(n) must also be true as well (-1)(-1)(-1)=(n)(n)(n)=1 -1=1 is untrue so (n)(n)(n) is false

As n squared is equal to 1, n must be either positive or negative 1. Positive 1 works with all three, and therefore doesn’t meet the problem’s parameters. However, negative 1 results in statements one and three being true, and statement two being false. Therefore, statement two is the correct answer.

N can equal any real number, lets say for simplicity n = -1. Think of this as in terms of powers: n X n = n^2, n X n X n = n^3, and n X n X n X n = n^4. We know that anything that is squared will come out positive and the same can be said for anything to the 4th power. For example -1^2 is really just -1 X -1 which equals 1 and -1^4 is really just -1 X -1 X -1 X -1 which equals 1. It is always true that anything to an even power will ALWAYS be positive . However anything to an odd power will always result in the number's original negative or positive sign . Take n^3 and realize that this is just saying n X n X n = n^3. Now lets take -1^3 which is just -1 X -1 X -1 = -1. Because a negative times a negative equals a positive and then times another negative equals a negative. This works for any odd power. -1^15 = -1 or -1^333333333 = -1. So therefor, saying that n X n X n = 1 given that n equals any real number is FALSE .

If n is √2 then n×n=2 But n×n×n=2√2 which is complex number

The simple wording of the problem is tricky. A more honest wording might read "When real number n belongs to a specific set, exactly one of the equations below is false:"

Of course to arrive to the solution, one has to assume that n belongs to the set of negative real numbers.

I didn't arrive to the correct answer based on negative numbers. I was honestly puzzled, so I picked the answer that wasn't like the other two: odd number of n 's vs. even number of n 's.

(-1)^3 isn’t 1.

n*n = n^2 n^2 = 1 n = +-√1 n= +-1

We are after a soltn where we have 1 incorrect soltn from the set, (1)^n always yields 1, while (-1)^n fluctuates between positive and negative.

let n be -1. Any negative number to an odd power results in a negative number. This applies with any negative number, as any negative number n raised to an odd power will be negative

Considering n to be -ve, equation 1 n^2 and equation 3 n^4 will be +ve hence n^3 should be -ve

If n=1/√(2)....... n²=1 ^ n⁴=1...... But n³=1/√(2) ....... Hence n³ is false. Is this an ok solution?

Any product of n an even numcer of times will always be > 0. For an odd number, it can be negative if n is negative. Ed Gray