This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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:

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.

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.

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.

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 $2^r m^2$.

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?)

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.

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.

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.

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.

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.

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.

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.

$(a-1)(a)(a+1)$ 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

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.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestHere'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.Log in to reply

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.

Log in to reply

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.

Log in to reply

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 $2^r m^2$.

Log in to reply

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

Log in to reply

Oh, of course, thanks for spotting that.

Log in to reply

I wasted almost 45 minutes doing a long and probably wrong solutions when it is so

** easy and elegant. Good work.Log in to reply

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

Log in to reply

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.

Log in to reply

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.

Log in to reply

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

Log in to reply

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.

Log in to reply

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.

Log in to reply

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.

Log in to reply

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.

Log in to reply

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.

Log in to reply

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

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

Log in to reply

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.

Log in to reply