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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, 2 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. \[ \]

Tim Vermeulen - 4 years, 2 months ago

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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.

Tim Vermeulen - 4 years, 2 months ago

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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.

Jimmy Kariznov - 4 years, 2 months ago

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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 \( 2^r m^2 \).

Calvin Lin Staff - 4 years, 2 months ago

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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

Ivan Stošić - 4 years, 2 months ago

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Oh, of course, thanks for spotting that.

Tim Vermeulen - 4 years, 2 months ago

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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

James Stevens - 4 years, 2 months ago

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

Jose Mateos - 4 years, 2 months ago

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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.

Brice Huang - 4 years, 2 months ago

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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.

Tim Vermeulen - 4 years, 2 months ago

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

Brice Huang - 4 years, 2 months ago

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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.

Andres Fabrega - 4 years, 2 months ago

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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.

Maurice Wei - 4 years, 2 months ago

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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.

Leonardo Cidrão - 4 years, 2 months ago

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Comment deleted Aug 02, 2013

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

Tim Vermeulen - 4 years, 2 months ago

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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.

Jose Mateos - 4 years, 2 months ago

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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.

Tim Vermeulen - 4 years, 2 months ago

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Comment deleted Aug 01, 2013

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

Tim Vermeulen - 4 years, 2 months ago

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

Jose Mateos - 4 years, 2 months ago

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

Tim Vermeulen - 4 years, 2 months ago

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

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

Snehdeep Arora - 4 years, 2 months ago

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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.

Jimmy Kariznov - 4 years, 2 months ago

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Comment deleted Aug 03, 2013

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

Tim Vermeulen - 4 years, 2 months ago

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