# No Positive Integer Solution

prove that there is no positive integer solutions of $a\left(a^2-1\right)=2b^2$ Note by Idham Muqoddas
6 years, 4 months ago

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Here's my second try, inspired by Jimmy and James:

$a$ and $a^2-1$ share no factors larger than $1$, so one of the two is a square, the other is twice a square. Obviously, $a^2-1$ is not a square and thus twice a square, which implies that $a$ is odd:

$a(a^2-1) = (a-1)a(a+1) = (2k-2)(2k-1)(2k) = 2b^2$

for some positive integer $k$. We know that $a=2k-1$ is a square, and thus both $2k-2$ and $2k$ are not squares, because there are no two positive squares with a difference of $1$. Also, $2k-2$ and $2k$ share no factors larger than $2$. But $(a+1)(a-1) = (2k-2)(2k)$ is twice a square, so one of the two factors must be a square, which is a contradiction. Therefore, there are no positive integers $a,b$ that satisfy the equation.

Note: I could also have said that $2k-2$ and $2k$ are both twice a square, so $k-1$ and $k$ are both squares, which is a contradiction. 

- 6 years, 4 months ago

Edit: this solution is wrong, I made a mistake.

$a(a^2-1) = (a-1)a(a+1),$

so we want to find three consecutive integers of which the product equals twice a square. This implies that one of the three factors is twice a square, the other two factors are squares. That means that we have to find two squares with a difference of $1$ or $2$. However, there are no two squares with a difference of $2$, so we have to look at two squares with a difference of $1$, and the only two squares with a difference of $1$ are $0$ and $1$. If one of the factors on the LHS is $0$, then $b=0$, but it had to be positive, so there are no positive integer solutions.

- 6 years, 4 months ago

Let $v_p(n)$ be the largest integer $k$ such that $p^k \mid n$.

Since $2b^2 = (a-1)a(a+1)$ we have $2v_p(b) + v_p(2) = v_p(a-1)+v_p(a)+v_p(a+1)$.

For $p \ge 3$, we have $v_p(2) = 0$ and at most one of $a-1,a,a+1$ is divisible by $p$.

Thus, $\{v_p(a-1), v_p(a), v_p(a+1)\} = \{2v_p(b),0,0\}$, i.e. one is $2v_p(b)$ and the other two are $0$.

Hence $v_p(a-1), v_p(a), v_p(a+1)$ are even for all $p \ge 3$.

Thus, $a-1, a, a+1$ are each of the form $2^r m^2$ for some non-negative integer $r$ and odd integer $m$.

Since $r$ is either even or odd, $a-1,a,a+1$ are each of the form $m^2$ or $2m^2$ for some integer $m$.

Now proceed as Tim V. has above, except also consider the case where all 3 are of the form $2m^2$, which is easily ruled out since 3 consecutive integers can't all be even.

- 6 years, 4 months ago

Great work. An identical way of expressing what you are saying is that for $a \geq 2$, since $\gcd(a, a-1) = 1, \gcd(a, a+1) = 1, \gcd(a-1, a+1) = 1$ or $2$, hence $a-1, a, a+1$ each must have the form $2^r m^2$.

Staff - 6 years, 4 months ago

Nope, that does not imply that one of these factors is a square. A product of non-squares can also be a square, for example 20*45 = 900

- 6 years, 4 months ago

Oh, of course, thanks for spotting that.

- 6 years, 4 months ago

I wasted almost 45 minutes doing a long and probably wrong solutions when it is so ** easy and elegant. Good work.

- 6 years, 4 months ago

Correct! You should add that if $a-1,a,$ or $a+1$ was not a square, then it would have an odd power of a prime factor. And, $b$ must have an even power of that factor, therefore one of the others (of $a-1,a,$ or $a+1$) must also have an odd power of that factor. This is not possible for prime factors $> 2$. (I know I explained that terribly. But if someone understood it could you please help clarify it?)

EDIT: Nevermind

- 6 years, 4 months ago

Suppose the equation has solutions. $\gcd (a, a^2-1) = 1$, so $a$ and $a^2-1$ must be, in some order, a square and twice a square. $a^2-1$ cannot be a square because it is between $(a-1)^2$ and $a^2$. Then $a$ must be the square. So, $a^2-1$ is twice a square. Let $a^2-1=(a+1)(a-1)=2k^2$. Clearly $a$ is odd, so $\frac{a+1}2$ and $\frac{a-1}2$ are integers. Furthermore $\gcd(\frac{a+1}2, \frac{a-1}2)=1$ because the two numbers differ by 1. As $\frac{a+1}2\cdot \frac{a-1}2= 2\left(\frac k2\right)^2$ is also twice a square, we have that $\frac{a+1}2$ and $\frac{a-1}2$ must be, in some order, a square and twice a square. Then $a+1$ and $a-1$ are twice a square and four times a square; thus they are also a square and twice a square (because four times a square is a square). However, then $a$ and one of $a-1$, $a+1$ are both squares. This is a contradiction.

- 6 years, 4 months ago

Nice one. You could add that it is a contradiction because the only two squares that differ by $1$ are $0$ and $1$, but those would result in a product of $0$ implying $b=0$, while $b$ had to be positive. But I guess that's quite straight forward and isn't really necessary.

- 6 years, 4 months ago

Thanks. Yes, I thought that was self-explanatory enough to omit.

- 6 years, 4 months ago

I've been trying a few things there and i think i have a solution (but i'm pretty sure it is wrong): We clear $b$, so it looks like: $\sqrt{ \frac {a}{2}}* \sqrt{(a^2 - 1)} = b$ Then, if $b$ has to be a positive integer, $\frac {\sqrt a} {\sqrt2} = \frac {n\sqrt2} {\sqrt2}$, so $a = 2n^2$, being $n$ a positive integer.. Hence, we have that $n\sqrt{a^2 - 1} = b$

If $a = 2n^2$, then $\sqrt {4n^4 - 1} = b$.

We can operate then so we have $8n^4 - 2 = 2b^2$, and then $a(a^2 - 1) = 8n^4 - 2$

We replace $a$ again so we have $2(4n^4 - 1) = 2n^2(4n^4-1)$.

So, $2 = 2n^2$. We replace $n$ and we have that $a = 2$, but it doesn't work because $2(4-1) \neq 2b^2$, so I have proved (or wasted my time with LaTeX because this is probably wrong) that there is no positive integer solutions of that equation.

- 6 years, 4 months ago

You say that

$b = \sqrt{ \frac{a}{2} } \cdot \sqrt{a^2-1}$

implies that $\sqrt{\frac{a}{2}}$ has to be an integer, but then you are assuming that the other factor, $\sqrt{a^2-1}$, is an integer as well, which it isn't. So there's your mistake. The rest is actually pretty nice.

- 6 years, 4 months ago

Define P(a)=(a-0)(a-1)(a+1) and F(b)=2(b-0)(b-0) as polynomials. The positive integer root of a is 1. Then we have: F(b)=P(1)=0 ---> F(b) doesn't have positive integer roots. Therefore, the given expression have no positive integer solutions.

- 6 years, 4 months ago

a and a^2-1 are coprime, thus they share no common factor p that divides b. Thus either p^2 divides a or a^2-1, for any prime divisor of b. note b>1, a must be positive integer solutions thus a^2-1>a. Then a^2-1 cant be a perfect square, so the factor expansion of a^2-1 = 2k^2, where k/b. However a^2-1 = (a-1)(a+1), so thus it can be split into two factors with gcd of 2. Therefore none of them can be perfect squares as a is a perfect square., contradiction. so thus no solutions for a.

- 6 years, 4 months ago

Easy problem. The left side = a (a+1) (a-1). Hence three consecutive integers. Then, we see two cases: Case 1: a even. That means the three consecutive integers are, in order, uneven, even,uneven. This three are obviously relatively prime integers, so it´s imposible to have all prime factors repeated 2n, since none of them are the same in the three. Case 2 : a uneven. That means the three consecutive integers are, in order, even, uneven,even. This means that only one factor is repeated twice, which is a two in both even numbers. Then, since the right side is multiplied by two, the right side will be divided by two, removing one of the 2´s. Therefore, only one two will appear in the prime descomposition of b^2. Since this is impossible(all factors must appear 2n times in the prime descomposition of a square), this is a contradiction, so there are no integer solutions. :) Nice problem.

- 6 years, 4 months ago

$a \times (a^2 - 1) = (a - 1) \times (a) \times (a+1) = 2b^2$

or $b = \sqrt {\frac {(a-1)(a)(a+1)} {2}}$

b has to positive integer or rational

$(a-1)(a)(a+1)$ might be a perfect square (rational number) but this rational number when divided by an irrational number ($\sqrt{2}$) will be an irrational number

$\Rightarrow$ b is irrational

$\Rightarrow$ there is no positive integer solution of the equation

there's my solution please tell me if i am wrong

- 6 years, 4 months ago

What if $(a-1)(a)(a+1)$ was twice a perfect square? Then, $\tfrac{(a-1)(a)(a+1)}{2}$ would be a perfect square, and so, $b = \sqrt{\tfrac{(a-1)(a)(a+1)}{2}}$ would be an integer, as required by the problem.

Your solution needs to justify that this cannot happen.

- 6 years, 4 months ago