# 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
4 years, 8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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. 

- 4 years, 8 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.

- 4 years, 8 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.

- 4 years, 8 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 - 4 years, 8 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

- 4 years, 8 months ago

Oh, of course, thanks for spotting that.

- 4 years, 8 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

- 4 years, 8 months ago

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

- 4 years, 8 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.

- 4 years, 8 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.

- 4 years, 8 months ago

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

- 4 years, 8 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.

- 4 years, 8 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.

- 4 years, 8 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.

- 4 years, 8 months ago

Comment deleted Aug 02, 2013

$$a \mid b^2$$ does not necessarily imply that $$a \mid b$$, e.g. $$(a,b) = (9,6)$$.

- 4 years, 8 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.

- 4 years, 8 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.

- 4 years, 8 months ago

Comment deleted Aug 01, 2013

That's not true, in fact, $$a(a^2-1)$$ is always even.

- 4 years, 8 months ago

You are right, but not all even numbers are equal to $$2b^2$$ if b is a positive integer.

- 4 years, 8 months ago

Right, but at least parity doesn't solve this problem immediately.

- 4 years, 8 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

- 4 years, 8 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.

- 4 years, 8 months ago

Comment deleted Aug 03, 2013

Don't forget to put \ ( and \ ) around your math expressions (without the spaces) in order to render them properly.

- 4 years, 8 months ago