No Positive Integer Solution

prove that there is no positive integer solutions of a(a21)=2b2a\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:

aa and a21a^2-1 share no factors larger than 11, so one of the two is a square, the other is twice a square. Obviously, a21a^2-1 is not a square and thus twice a square, which implies that aa is odd:

a(a21)=(a1)a(a+1)=(2k2)(2k1)(2k)=2b2 a(a^2-1) = (a-1)a(a+1) = (2k-2)(2k-1)(2k) = 2b^2

for some positive integer kk. We know that a=2k1a=2k-1 is a square, and thus both 2k22k-2 and 2k2k are not squares, because there are no two positive squares with a difference of 11. Also, 2k22k-2 and 2k2k share no factors larger than 22. But (a+1)(a1)=(2k2)(2k)(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,ba,b that satisfy the equation.


Note: I could also have said that 2k22k-2 and 2k2k are both twice a square, so k1k-1 and kk are both squares, which is a contradiction.

Tim Vermeulen - 6 years, 4 months ago

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


a(a21)=(a1)a(a+1),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 11 or 22. However, there are no two squares with a difference of 22, so we have to look at two squares with a difference of 11, and the only two squares with a difference of 11 are 00 and 11. If one of the factors on the LHS is 00, then b=0b=0, but it had to be positive, so there are no positive integer solutions.

Tim Vermeulen - 6 years, 4 months ago

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Let vp(n)v_p(n) be the largest integer kk such that pknp^k \mid n.

Since 2b2=(a1)a(a+1)2b^2 = (a-1)a(a+1) we have 2vp(b)+vp(2)=vp(a1)+vp(a)+vp(a+1)2v_p(b) + v_p(2) = v_p(a-1)+v_p(a)+v_p(a+1).

For p3p \ge 3, we have vp(2)=0v_p(2) = 0 and at most one of a1,a,a+1a-1,a,a+1 is divisible by pp.

Thus, {vp(a1),vp(a),vp(a+1)}={2vp(b),0,0}\{v_p(a-1), v_p(a), v_p(a+1)\} = \{2v_p(b),0,0\}, i.e. one is 2vp(b)2v_p(b) and the other two are 00.

Hence vp(a1),vp(a),vp(a+1)v_p(a-1), v_p(a), v_p(a+1) are even for all p3p \ge 3.

Thus, a1,a,a+1a-1, a, a+1 are each of the form 2rm22^r m^2 for some non-negative integer rr and odd integer mm.

Since rr is either even or odd, a1,a,a+1a-1,a,a+1 are each of the form m2m^2 or 2m22m^2 for some integer mm.

Now proceed as Tim V. has above, except also consider the case where all 3 are of the form 2m22m^2, which is easily ruled out since 3 consecutive integers can't all be even.

Jimmy Kariznov - 6 years, 4 months ago

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Great work. An identical way of expressing what you are saying is that for a2 a \geq 2, since gcd(a,a1)=1,gcd(a,a+1)=1,gcd(a1,a+1)=1 \gcd(a, a-1) = 1, \gcd(a, a+1) = 1, \gcd(a-1, a+1) = 1 or 22, hence a1,a,a+1 a-1, a, a+1 each must have the form 2rm2 2^r m^2 .

Calvin Lin Staff - 6 years, 4 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 - 6 years, 4 months ago

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Oh, of course, thanks for spotting that.

Tim Vermeulen - 6 years, 4 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 - 6 years, 4 months ago

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Correct! You should add that if a1,a,a-1,a, or a+1a+1 was not a square, then it would have an odd power of a prime factor. And, bb must have an even power of that factor, therefore one of the others (of a1,a,a-1,a, or a+1a+1) must also have an odd power of that factor. This is not possible for prime factors >2> 2. (I know I explained that terribly. But if someone understood it could you please help clarify it?)

EDIT: Nevermind

James Stevens - 6 years, 4 months ago

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Suppose the equation has solutions. gcd(a,a21)=1\gcd (a, a^2-1) = 1, so aa and a21a^2-1 must be, in some order, a square and twice a square. a21a^2-1 cannot be a square because it is between (a1)2(a-1)^2 and a2a^2. Then aa must be the square. So, a21a^2-1 is twice a square. Let a21=(a+1)(a1)=2k2a^2-1=(a+1)(a-1)=2k^2. Clearly aa is odd, so a+12\frac{a+1}2 and a12\frac{a-1}2 are integers. Furthermore gcd(a+12,a12)=1\gcd(\frac{a+1}2, \frac{a-1}2)=1 because the two numbers differ by 1. As a+12a12=2(k2)2\frac{a+1}2\cdot \frac{a-1}2= 2\left(\frac k2\right)^2 is also twice a square, we have that a+12\frac{a+1}2 and a12\frac{a-1}2 must be, in some order, a square and twice a square. Then a+1a+1 and a1a-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 aa and one of a1a-1, a+1a+1 are both squares. This is a contradiction.

Brice Huang - 6 years, 4 months ago

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Nice one. You could add that it is a contradiction because the only two squares that differ by 11 are 00 and 11, but those would result in a product of 00 implying b=0b=0, while bb had to be positive. But I guess that's quite straight forward and isn't really necessary.

Tim Vermeulen - 6 years, 4 months ago

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

Brice Huang - 6 years, 4 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 bb, so it looks like: a2(a21)=b \sqrt{ \frac {a}{2}}* \sqrt{(a^2 - 1)} = b Then, if bb has to be a positive integer, a2=n22 \frac {\sqrt a} {\sqrt2} = \frac {n\sqrt2} {\sqrt2}, so a=2n2 a = 2n^2, being nn a positive integer.. Hence, we have that na21=b n\sqrt{a^2 - 1} = b

If a=2n2 a = 2n^2, then 4n41=b \sqrt {4n^4 - 1} = b .

We can operate then so we have 8n42=2b2 8n^4 - 2 = 2b^2 , and then a(a21)=8n42 a(a^2 - 1) = 8n^4 - 2

We replace aa again so we have 2(4n41)=2n2(4n41) 2(4n^4 - 1) = 2n^2(4n^4-1) .

So, 2=2n22 = 2n^2. We replace nn and we have that a=2 a = 2 , but it doesn't work because 2(41)2b2 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 - 6 years, 4 months ago

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You say that

b=a2a21 b = \sqrt{ \frac{a}{2} } \cdot \sqrt{a^2-1}

implies that a2\sqrt{\frac{a}{2}} has to be an integer, but then you are assuming that the other factor, a21\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 - 6 years, 4 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 - 6 years, 4 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 - 6 years, 4 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 - 6 years, 4 months ago

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a×(a21)=(a1)×(a)×(a+1)=2b2a \times (a^2 - 1) = (a - 1) \times (a) \times (a+1) = 2b^2

or b=(a1)(a)(a+1)2b = \sqrt {\frac {(a-1)(a)(a+1)} {2}}

b has to positive integer or rational

(a1)(a)(a+1)(a-1)(a)(a+1) might be a perfect square (rational number) but this rational number when divided by an irrational number (2\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 - 6 years, 4 months ago

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What if (a1)(a)(a+1)(a-1)(a)(a+1) was twice a perfect square? Then, (a1)(a)(a+1)2\tfrac{(a-1)(a)(a+1)}{2} would be a perfect square, and so, b=(a1)(a)(a+1)2b = \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 - 6 years, 4 months ago

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