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How to Find all primes $p$ such that $\dfrac{(11^{p-1}-1)}{p}$ is a perfect square

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If $p \neq 2,5$ then 10 divides the top and hence 100 must divide it too...Now, 100 divide $11^{p-1}-1$ if and only if 10 divides p-1... But here I am stuked... –  N. S. Jul 12 '12 at 13:47
    
@N.S. "stuck".${}{}{}$ –  Pedro Tamaroff Jul 15 '12 at 4:23
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4 Answers

up vote 4 down vote accepted

Here is another idea. I'm going to argue that there is no such prime.

I'll make use of a technique called lifting the exponent. This will serve primarily to shorten several steps of my argument (i.e., it is not completely essential). In brief, the method requires the following result, which is not terribly difficult to verify. If $p > 2$ is prime and $a$ and $b$ are integers with $v_p(a) = v_p(b) = 0$ and $v_p(a - b) > 0$, then for any positive integer $n$, \begin{align*} v_p(a^n - b^n) = v_p(a - b) + v_p(n). \end{align*} The assertion is also valid if $p = 2$, provided that $v_2(a-b) > 1$, i.e., provided that $4 \mid a-b$. Note that a nonzero integer $m$ is a perfect square if and only if $v_p(m) \equiv 0 \pmod2$ for all primes $p$.

I will also use the fact that the set of quadratic residues modulo $11$ is $\{1,3,4,5,9\}$, and therefore that the set of nonresidues is $\{2,6,7,8,10\}$. In particular, $-1,2$, and $7$ are quadratic nonresidues modulo $11$. Note that, if $p$ is a prime such that $(11^{p-1} - 1)/p$ is a perfect square, then the residue class $\overline{p}$ of $p$ modulo $11$ is a quadratic nonresidue. Finally, one can check directly that $(11^{p-1} - 1)/p$ is not a perfect square when $p = 2,3,5$, or $7$.


Here is the argument.

  1. Assume for a contradiction that $(11^{p-1} - 1)/p$ is a perfect square for some prime $p$. First of all $2 \mid p-1$, so $2^3\cdot 3\cdot 5 = 11^2 - 1 \mid 11^{p-1} - 1$. For $q = 2,3,5$, one has $$ v_q(11^{p-1} - 1) = v_q(11^2 - 1) + v_q\left(\frac{p-1}{2}\right) \equiv 1 + v_q\left(\frac{p-1}{2}\right) \pmod 2. $$ The left-hand side is even by assumption (since $p > 7$), so $v_q\left(\frac{p-1}{2}\right)$ is odd. This proves that $3$ and $4$ divide $p-1$. One can check that $v_7(11^3 - 1) = 1$. It follows that $$ v_7(11^{p-1} - 1) = 1 + v_7\left(\frac{p-1}{3}\right). $$ The left-hand side is even (again, since $p > 7$), so $r = v_7\left(\frac{p-1}{3}\right)$ is odd.

  2. Now write $p - 1 = 12 \cdot 7^r m$, so that $7\nmid m$. We claim that $11^{12m} - 1 = 7a^2$ for some integer $a$, with $p\nmid a$. The crucial part is that $p\nmid 11^{12m} - 1$. Since $7 \nmid 4m$ and $v_7(11^3 - 1)>0$, we can lift the exponent to obtain $$ v_7(11^{12m} - 1) = v_7(11^3 - 1) + v_7(4m) = v_7(11^3 - 1) = 1. $$ Assume now that $q\not = 7$ is any prime divisor of $11^{12m} - 1$. Then we can lift the exponent again to get $$ v_q(11^{12m} - 1) = v_q(11^{p-1} - 1) - v_q(7^r) = v_q(11^{p-1} - 1). $$ (When $q = 2$ we use the fact that $4 \mid 11^{12m} - 1$.) The right-hand side is even unless possibly $q = p$. So we have $11^{12m} - 1 = 7p^ea^2$ where $a$ is an integer not divisible by $p$ and $e$ is either $0$ or equal to $v_p(11^{p-1} - 1)$, in which case it is odd. The residue $\overline{7p^ea^2} = -1$ is a quadratic nonresidue modulo $11$. Since $\overline{p}$ and $7$ are both quadratic nonresidues and $\overline{a^2}$ is a quadratic residue, it must be the case that $e = 0$. (A product of quadratic nonresidues is a quadratic residue, likewise a product of quadratic residues is a quadratic residue.) Thus $11^{12m} - 1 = 7a^2$, with $p \nmid a$, as claimed.

  3. Ok, just one more step. Basically, we use an argument similar to the one above to show that $11^{6m} - 1 = 14b^2$ for some integer $b$. Since $\overline{14} = 3$ and $\overline{b^2}$ are quadratic residues modulo $11$, the product $\overline{14b^2} = -1$ will be as well, and this then gives the desired contradiction. First, since $11^{6m} - 1 \mid 11^{12m} - 1$ and $p \nmid 11^{12m} - 1$, we know that $p \nmid 11^{6m} - 1$. Thus if $q$ is a prime that divides $11^{6m} - 1$, then $q \not = p$, and by lifting the exponent we get $$ v_q(11^{6m} - 1) = v_q(11^{p-1} - 1) - v_q(2\cdot 7^r) \equiv v_q(2\cdot 7^r) \pmod 2. \tag{1} $$ If $q \not = 2$ or $7$, then $v_q(2) = v_q(7^r) = 0$, so $v_q(11^{6m} - 1)$ is even. Since $4 \mid 11^{6m} - 1$ and $7\mid 11^{6m} - 1$, the equation $(1)$ shows that $$v_2(11^{6m} - 1)\equiv 1\pmod 2\quad \text{and} \quad v_7(11^{6m} - 1) \equiv r \equiv 1\pmod 2,$$ i.e., both numbers are odd. (Recall that $r = v_7\left( \frac{p-1}{3}\right)$ is odd, as was shown in 1.) We conclude that $11^{6m} - 1 = 14 b^2$ for some integer $b$, as claimed, and we derive a contradiction as indicated at the beginning of this step.

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Good answer Strehlke! –  Daniel Montealegre Jul 18 '12 at 21:12
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Let $\frac{a^{p-1}-1}{p}=b^2$ for some integer b.

If p=2, $a=2b^2+1$(11 is not of the form).

If p>2 & prime ,so must be odd=(2k+1) (say)

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

Now, if a is odd (like 11), $a^k±1$ is even => b is even =2d(say).

$\frac{a^k+1}{2}\frac{a^k-1}{2}=d^2(2k+1)$

Now, $(\frac{a^k+1}{2},\frac{a^k-1}{2})$=1

So, either $\frac{a^k+1}{2}=m^2p$ and $\frac{a^k-1}{2}=n^2$ or vice versa.

Now, $n^2$≡0,1,4,9,5,3(mod 11) =>$11∤(2n^2-1)$

In other case,

If $11^k=2n^2+1$

n must be even=2m (say)=> $2n^2+1≡1(mod\ 8)$.

Now, 11≡3(mod 8) =>$11^2≡1(mod\ 8)$=>$11^{2r}≡1(mod\ 8)$ and $11^{2r+1}≡3(mod\ 8)$

So, k must even=2r(say).

The part problem becomes $(11^r)^2-2n^2=1$ which is well known Pell equation.

The minimum solution in natural number for $A^2-2B^2=1$ is (3,2).

So, A is $\sum_{s≥2t≥0}sC_{2t}3^{s-2t}(2\sqrt2)^{2t}$ where s is a natural number, can this be a power of 11? Find here

An observation:

$a^k≡-1(mod\ p)=>a^{2k}≡1(mod\ p) $

Let $d=ord_pa$ => d|2k,but d∤k and d∤2 as k is even

=> a(=11) must be a primitive root of p to admit any solution.

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So is there a prime p such that (11^(p-1) - 1)/p is a perfect square? –  Frank Jul 13 '12 at 3:11
    
I missed to mention, the last condition in necessary, yet to prove sufficiency. –  lab bhattacharjee Jul 13 '12 at 4:45
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If $p=2$, $\frac{11^{p-1}-1}{p}$ is not a perfect square. In the following, we assume that $p\neq 2$.

Suppose that $\frac{11^{p-1}-1}{p}$ is a perfect square and write $p=2k+1$. $$11^{2k}-1 = (11^k-1)(11^k+1)= p n^2 $$ Now, as $gcd(11^k-1, 11^k+1)=2$, we have one of the following cases $$ 11^k-1 = p a^2\text{ and }11^k +1 = b^2 $$

$$ 11^k-1 = a^2\text{ and }11^k +1 =p b^2 $$

$$ 11^k-1 = 2 p a^2\text{ and }11^k +1 = 2 b^2 $$

$$ 11^k-1 = 2 a^2\text{ and }11^k +1 =2 p b^2 $$

In the first case, $11^k=(b-1)(b+1)$ which is impossible.

In the second case, $11^k-1 = a^2$ so $a$ is even and $4$ divides $a^2$. But $a^2=pb^2-2$ so, $pb^2=2$ modulo $4$, and $b^2=2$ modulo $4$, which is impossible.

In the third case, $2 b^2 = 1$ modulo $11$ which is impossible, as it is easy to verify that the squares modulo $11$ are 0, 1, 4, 9, 5, 3.

The fourth case remains to be examined.

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Not sure about your argument that $11^k-1$ and $11^k+1$ being coprime - are they not both even? Having said that, I am sure your argument can be rescued following ideas similar to this question: math.stackexchange.com/questions/161599/… –  Old John Jul 12 '12 at 15:40
    
You are right, but as $p\neq2$, it is easy to caorrect the argument. –  saposcat Jul 12 '12 at 15:45
    
Is the fourth case impossible either ? –  Frank Jul 12 '12 at 17:01
    
I don't know actually how to deal with the fourth case. –  saposcat Jul 12 '12 at 21:55
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In the solution proposed by saposcat, the fourth case remained open. Here is the continuation of the solution in that case.

Let we have $11^k−1=2a^2$ and $ 11^k+1=2pb^2.$ If $3\not|a$, then $11^k≡0(mod \ 3)$, which is not possible, so $3|a$. Therefore $11^k≡1(mod\ 3)$, that's why k should be even. Let's suppose $k=2l$. Then we have $11^{2l}-1 = 2a^2$, so $(11^l-1)(11^l+1) = 2a^2$. We get that $11^l-1 = 2m^2$ and $11^l+1 = n^2$ or $11^l-1 = m^2$ and $11^l+1 = 2n^2$, for both cases saposcat got contradictions (like 1st and 3rd cases).

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