I cannot do it without Gauss reciprocity law. Btw, is it admissible to use reciprocity law at IMO?

meandre wrote: I cannot do it without Gauss reciprocity law. Btw, is it admissible to use reciprocity law at IMO? Yes. For example in the Iberoamerican of 2003, the sixth problem can be solve for Legendre's symbol and law of reciprocity quadratic.

I think there is no restriction is using any formula or law

Using quadratic residues and quadratic reciprocity, I have a solution as long as $ab$ is not a perfect square, but if $a,b$ are distinct integers whose product IS a perfect square (ie $a=kc^2$ and $b=kd^2$ with $c\neq d$), then the whole thing breaks down. The idea is

meandre wrote: I cannot do it without Gauss reciprocity law. Btw, is it admissible to use reciprocity law at IMO? Certain concepts and results, such as Dirichlet's theorem or quadratic reciprocity are admissible in the students' solutions, but I would assume that many leaders would reject a problem that NEEDS these concepts in order to solve it, on the grounds that not all participating countries have these concepts on the high-school programs. I guess that the general idea is, you can use them in your solution, but it is highly preferred (read "mandatory") that any contest problem (except for maybe 3 and/or 6) is solvable without the need of such results. This is my understanding of the situation, I may be wrong... Now again, if my understanding is not entirely off mark, and no solutions in the shortlist were provided for this problem without the use of quadratic reciprocity, then my bet is that this was one of the first shorlisted problems to be scratched by the leaders, on the grounds that I mention; or maybe Dirichlet's theorem is out, and quadratic reciprocity is in, and that's why it asks especifically for one value of $n$, and not infinitely many...

We will show that if $(a^n-1)(b^n-1)$ is a perfect square for all $n$, then $a=b$. Lemma 1: The sets of prime divisors of $a^n - 1$ and $b^n - 1$ are equal for all positive integers $n$. Proof: By symmetry, it suffices to show that if $p | a^n - 1$, then $p | b^n - 1$. Suppose for the sake of contradiction that $p | a^n - 1$ but $p \not | b^n - 1$. Then $\nu_p(a^n - 1)$ must be even, since $(a^n - 1)(b^n - 1)$ is a perfect square. But $p \not | b^n - 1$ implies $p \not | b^{np} - 1$, whence $\nu_p(a^{pn} - 1) = 1 + \nu_p(a^n - 1)$ must be odd, so $(a^{pn} - 1)(b^{pn} - 1)$ cannot be a perfect square. Lemma 2: Let $x_1, x_2, \ldots, x_n$ be distinct rationals, and let $c_1, c_2, \ldots, c_n$ be real numbers. Then if there exists a sequence of integers $s_1, s_2, \ldots$ so that $\lim_{k \to \infty} [ s_k - (c_1 x_1^k + c_2 x_2^k + \ldots + c_n x_n^k)] = 0$, then each $x_i$ with absolute value not less than 1 must be an integer. Proof: It is clear that it is sufficient to prove this when $|x_i| \geq 1$ for all $i$. Let $q_k = c_1 x_1^k + c_2 x_2^k + \ldots + c_n x_n^k$, and let $d_k = s_k - q_k$. We induct on $n$. When $n = 1$, let $x = \frac{p}{q}$. Then \begin{align*} qs_{k+1} - ps_k &= q\left (d_{k+1} + c_1 \frac{p^{k+1}}{q^{k+1}} - p \left (d_k + c_1 \frac{p^k}{q^k} \right) \right) \\ &= q d_{k+1} - p d_k. \end{align*} $\lim_{k \to \infty} (qd_{k+1} - p d_k) = 0$, so $\lim_{n \to \infty} (qs_{k+1} - ps_k) = 0$. But $qs_{k+1} - ps_k$ is always an integer, so we find that $qs_{k+1} = p s_k$ for all sufficiently large $k$. Hence, for some $k$ and all nonnegative $i$, $s_{k+i} = \left(\frac{p}{q}\right)^i s_k$, But $s_{k+i}$ is always a nonzero integer for $k$ sufficiently large (since $c_i x_1^k$ is unbounded and $s_k - c_i x_1^k$ converges), so we must have $q=1$. We will now show that if the result is true for some $n$, then it is also true for $n+1$. Let $x_{n+1} = \frac{p_{n+1}}{q_{n+1}}$. Then \begin{align*} q_{n+1} s_{k+1} - p_{n+1} s_k &= q_{n+1} \left (d_{k+1} + c_{n+1} \frac{p_{n+1}^{k+1}}{q_{n+1}^{k+1}} + \sum_{i=1}^n c_i x_i^{k+1} \right) - p_{n+1} \left (d_k + c_{n+1} \frac{p_{n+1}^k}{q_{n+1}^k} + \sum_{i=1}^n c_i x_i^k \right) \\ &= (q_{n+1} d_{k+1} - p_{n+1} d_k) + \sum_{i=1}^n (q_{n+1} x_i c_i - p_{n+1} c_i) x_i^k. \end{align*} Hence, \[ \lim_{k \to \infty} \left [ (q_{n+1} s_{k+1} - p_{n+1} s_k ) -\sum_{i=1}^n (q_{n+1} x_i c_i - p_{n+1} c_i) x_i^k \right ] = \lim_{k \to \infty} \left [ q_{n+1} d_{k+1} - p_{n+1} d_k \right] = 0,\] so if we define $s_k' = q_{n+1} s_{k+1} - p_{n+1} s_k$ and $c_i' = q_{n+1} x_i c_i - p_{n+1} c_i$, we may apply the inductive hypothesis to find that each $x_i$, $1 \leq i \leq n$, is an integer. By symmetry (e.g., by isolating $x_1$ instead of $x_{n+1}$), we may conclude that $x_{n+1}$ is an integer as well. $\qed$ Let $n$ be any integer. Take a positive integer $p$ so that $p > 3n$ and so that the reals $r_1 = \frac{(ab)^n}{a^p}$ and $r_2 = \frac{(ab)^n}{b^p}$ are both less than 1. Write the Taylor series of $\sqrt{1-x^n}$ about 0 as $\sum_{i=0}^{p-1} c_i x^i + O(x^p)$, for some reals $c_0, c_1, c_2, \ldots, c_{p-1}$. Consider the sequence given by $s_k = a^{nk} b^{nk} \left (\sum_{i=0}^{p-1} c_i a^{-i} \right ) \left (\sum_{i=0}^{p-1} c_i b^{-i} \right )$. We have \begin{align*} \sqrt{(a^{nk} - 1)(b^{nk} - 1)} - s_k &= a^{nk} b^{nk} \sqrt{(1 - a^{-nk})(1 - b^{-nk})} - s_k \\ &= a^{nk} b^{nk} \left( \sum_{i=0}^{p-1} c_i a^{-ik} + O(a^{-kp}) \right) \left( \sum_{i=0}^{p-1} c_i b^{-ik} + O(b^{-kp}) \right) - a^{nk} b^{nk} \left( \sum_{i=0}^{p-1} c_i a^{-ik} \right) \left( \sum_{i=0}^{p-1} c_i b^{-ik} \right) \\ &= a^{nk} b^{nk} \left( O(b^{-kp}) \left( \sum_{i=0}^{p-1} c_i a^{-ik} \right) + O(a^{-kp}) \left( \sum_{i=0}^{p-1} c_i a^{-ik} \right)+ O(a^{-kp} b^{-kp})\right) \\ &= O(r_1^k) O(1) + O(r_2^k) O(1) + O(r_1^k b^{-kp}). \end{align*} Hence, $\lim_{k \to \infty} {\sqrt{(a^{nk} - 1)(b^{nk}-1)} - s_k = 0}$. Furthermore, by expanding $a^{nk} b^{nk} \left (\sum_{i=0}^{p-1} c_i a^{-i} \right ) \left (\sum_{i=0}^{p-1} c_i b^{-i} \right )$, we find that $s_k$ is a linear combination of terms of the form $\left(a^{t_1} b^{t_2} \right)^k$, where $n - p \leq t_1, t_2 \leq n$ are integers. Since $a^{t_1} b^{t_2}$ is always rational and $\sqrt{(a^{nk} - 1)(b^{nk}-1)}$ is always an integer, lemma 4 tells us that any number of the form $\frac{a^{e_1}}{b^{e_2}}$ not less than 1, with $|e_1|, |e_2| < n$, must be an integer. By taking $n$ to be sufficiently large, we find that all rationals of the form $\frac{a^{e_1}}{b^{e_2}}$ that are not less than 1 must be integers. We now claim that $\alpha = \log_b a$ is rational. Let $\frac{u}{v}$ be a convergent in the continued fraction expansion of $\alpha$, with $\gcd(u,v) = 1$ and $0 \leq u \alpha - v < \log_b 2$. Then $1 \leq \frac{a^u}{b^v} = b^{u \alpha - v} < b^{\log_b 2} = 2$. But $\frac{a^u}{b^v}$ is an integer, so $a^u = b^v$, so $\alpha = \frac{v}{u}$ is rational. For all primes $p$, $u v_p(a) = v v_p(b)$. because $\gcd(u,v) = 1$, $u | v_p(b)$ and $v | v_p(a)$, whence there exists some integer $c$ such that $a = c^v$ and $b = c^u$. Let $u' = 7u$ and $v' = 7v$. By lemma 1, the set of prime factors of $c^{u'} - 1$ and $c^{v'} - 1$ must be the same. Suppose without loss of generality that $u' \leq v'$. By Zsigmondy's theorem, we can find some prime $p$ such that the order of $c$ modulo $p$ is $v'$ (we avoid the exceptional cases by noting that $u', v' > 6$, and that $c \neq 1$ since $a,b \neq 1$.) If $u' < v'$, then $p \not | c^{u'} - 1$, which contradicts lemma 1. Hence, $u' = v'$, so $a=b$, as desired.

I'd like to add that $(a^4-1)(b^4-1)$ is basically never a square for distinct $a,b>1$. Indeed, this implies that $a^4-1=dm^2$ and $b^4-1=dn^2$ for some positive integers $m,n$ with $d$ square-free WLOG. By this paper (which contains a fairly short quadratic reciprocity proof), we see that $d=1785$, and WLOG $(a,b)=(13,239)$, for which \[(13^4-1)(239^4-1)=9653280^2.\]But \[(13-1)(239-1)=2856\]is not a perfect square. By the way, this is a stronger problem pythag011 came up with (see my second post in that thread). His motivation was probably that this problem requires $(a^2-1)(b^2-1),a^2b^2,(a^2+1)(b^2+1)$ to be perfect squares (the last because $(a^4-1)(b^4-1)$ is a square).

If one could find a nontrivial bound for $$\sum_{n\pmod{k}}\left(\ \dfrac{(a^n-1)(b^n-1)}{k} \right)$$ where the $\left(\frac{}{}\right)$ denotes the Jacobi Symbol, then I can solve (a stronger version of) this problem. Any ideas for that sum?

Can someone explain more clearly Lemma 2 of Zhero's solution and how it helps to solve the problem? thanks

For n=$2t$ we have $($(ab)^t$-1)^2-$ ($a^t$-$b^t$)^2 $=$c^2 $. a$!=$b so (ab)^t-1, a^t$-$b^t$ and c are Pithagorean triple, so (ab)^t-1=x^2+y^2 for some x,y. So it is enough to prove that for every $k=ab$ there exist natural t for which p=4s+3, p-prime divides $k^t-1$ (Because then $x^2$+$y^2& is not divisible with that p) We may simply put t=q-1, for some q-prime, q=4m+3, and k is not divisible with q.

Your claim is valid iff the Pythagorean triple is primitive but that does not holds if $(a-1,b-1)>1$

Does anyone know how to prove the following claim with elementary number theory method? "Given positive integers $a,b$. If for all prime number $p\nmid ab$,$ord_p(a)=ord_p(b)$, then $a=b$." It is just the same thing as Lemma1 in Zhero's solution. I think that it is right but I don't have any idea to prove it.><

The above user's claim follows from a theorem due to Schinzel, which kills the problem: one concludes that $a$ is a power of $b$ and vice versa, hence $a=b$.

Oh wow... trivial-looking NT can get quite deep at times... Of course, we assume to the contrary- First things first, playing around with $v_p(a-1)$ and $v_p(b-1)$, we get that if $p \mid a-1$, then $p \mid b-1$. Why? Say $p \mid a-1$, but $b \not \equiv 1 \pmod{p}$. If $v_p(a-1)$ is odd, we're done. If $v_p(a-1)$ is even, then consider $n=p$. Clearly, $v_p(a^p-1)=v_p(a-1) +1$, which is odd. But $b^p-1 \equiv b-1 \not \equiv 0 \pmod{p}$, thus $v_p((a^p-1)(b^p-1))$ is odd $=>$Contradiction. Thus $p \mid a^n-1 <=> p \mid b^n-1$. Hence, clearly, we need $O_p(a)=O_p(b)$ for all primes that don't divide $a$ or $b$. But this clearly implies $a=b$ as, for WLOG $a>b$, $\frac{a^l-1}{b^l-1}$ can get arbitrarily large, thus we're done.

Short proof, but certainly took a while to find... Great problem!

@Above: you proved that $p$ is a prime divisor of $a-1$ if and only if $p$ is a prime divisor of $b-1$, but you use the fact that $p | a^n - 1 \Leftrightarrow p | b^n - 1$. How does this follow from the case $n = 1$? @Below. Thanks. EDIT: in fact, a more serious problem is proving that if $ord_p(a) = ord_p(b)$ for all $p$, then $a = b$. This problem was already noted in posts 14 and 15.

Loppukilpailija wrote: @Above: you proved that $p$ is a prime divisor of $a-1$ if and only if $p$ is a prime divisor of $b-1$, but you use the fact that $p | a^n - 1 \Leftrightarrow p | b^n - 1$. How does this follow from the case $n = 1$? Simply take $a \leftarrow a^n, b \leftarrow b^n$. Also note that this is not a consequence of the case $n=1$ but is indeed independent of it.

We provide a solution to this problem using non-elementary methods, namely Frobenius density theorem / Chebotarev's density theorem. The solution is motivated by investigations on problems on quadratic residues. The post was written quite hastily. The letter $p$ will always denote a prime number. The problem "Let $a$ be an integer. Assume that $a$ is a quadratic residue modulo all large enough $p$. Prove that $a$ is the square of some integer." is a classic one. The canonical proof of the claim is by the multiplicativity of Legendre symbol, the quadratic reciprocity, and finally we use Dirichlet's theorem (and Chinese remainder theorem). The variant "Let $a$ be an integer. Assume that $a$ is a cubic residue modulo all large enough $p$. Prove that $a$ is the cube of some integer." does not seem to be solvable by a similar procedure. (Though a cubic reciprocity law does exist, it is not as satisfying as one would hope.) Another example: "Let $a_1, a_2, \ldots , a_n$ be (non-zero) integers. Prove that there exists infinitely many $p$ such that all of $a_i$ are quadratic residues modulo $p$." This can be solved as the classic problem above. However, the variant for higher power residues do not follow by a similar proof. In this case one can prove the statement by proving that in fact for any $n$ non-constant polynomials with integer coefficients, the set of their common prime divisors (i.e. the set of primes which divide some value of any of the polynomials at some integer point) is infinite. This problem is solvable using algebraic number theory by reducing to the case $n = 1$, as has been discussed on the forum a couple of times. One more variant: "For which integers $a_1, \ldots , a_n$ are there infinitely many $p$ such that all $a_i$ are quadratic non-residues modulo $p$?". Again, this is solvable, but does not generalize for higher powers. The above mentioned fact about the infinitude of prime divisors of polynomials does not help either. The quadratic reciprocity argument is kind of ad hoc in the sense that it does not generalize to higher powers. One would therefore hope that there exists some method which does generalize. There exists such a method, the Frobenius density theorem, or the more general Chebotarev density theorem. Since this is a broad topic, our introduction is quite brief. Fist, we quickly present the theorem, then solve a generalization of the classic problem, and finally solve the IMO SL 2009 N7 as an application. First problem. Assume $a$ is a cubic residue for all large enough $p$. Prove that $a$ is the cube of some integer. Solution. Consider the polynomial $P(x) = x^3 - a$. We are given that $P(x) \equiv 0 \pmod{p}$ is solvable for almost all $p$. The Frobenius density theorem provides a way to inspect this set of primes. We now introduce the theorem. The rational numbers $\mathbb{Q}$ form a field, which roughly means that all operations are well-defined and behave as expected. In particular, one can divide by non-zero elements (compare to $\mathbb{Z}$). One can form fields by extending a given field. For example, the complex numbers $\mathbb{C}$ are formed by adding the element $i$ with $i^2 = -1$ to the field $\mathbb{R}$ of real numbers. Similarly, one can extend the rational numbers. For example, by extending $\mathbb{Q}$ with the element $i$ (denoted $\mathbb{Q}(i)$) we get the set of numbers of the form $a + bi$ with $a$ and $b$ rational. One may check that this indeed is a field. In general, one can extend the rational numbers with some algebraic number $\alpha$. The resulting field will contain numbers of the form $q_0 + q_1\alpha + \ldots + q_{n-1}\alpha^{n-1}$, where $n$ is the degree of the minimal polynomial of $\alpha$ (which means that $\alpha^n$ may be expressed in terms of $\alpha^i$ for $i < n$). It's not trivial that the quotients of such numbers is of this form. The proof is omitted. One may further extend $\mathbb{Q}(\alpha)$ by $\beta$. This is denoted simply by $\mathbb{Q}(\alpha, \beta)$. This field, too, simply consists of polynomials of the numbers $\alpha$ and $\beta$. Let $\alpha = \sqrt[3]{a}$, and let $\zeta_3$ be a primitive third root of unity, that is, $\zeta_3^3 = 1$ and $\zeta_3^k \neq 1$ for all $1 \le k < 3$. Now, the roots of $P$ are $\alpha, \alpha\zeta_3$, and $\alpha\zeta_3^2$. We see that all of the roots belong to the field $K = \mathbb{Q}(\alpha, \zeta_3)$. Furthermore, this is the smallest extension of $\mathbb{Q}$ which contains all roots of $P$. We call the field with this property the splitting field of $P$. We have just one thing to define before introducing the density theorem. We know that in $\mathbb{C}$ the map $f$ with $f(a+bi) = a - bi$, the so-called complex conjugation, is an isomorphism of $\mathbb{C}$ to itself fixing $\mathbb{R}$. In other words, it is bijective, additive, multiplicative, and $f(x) = x$ for all $x \in \mathbb{R}$. There's also another such isomorphism, the trivial one $g(a+bi) = a+bi$ for all $a, b \in \mathbb{R}$. However, these are the only ones: for any such map $h$ we must have $h(i)^2 = h(i^2) = h(-1) = -1$, so $h(i)$ is $\pm i$. How does this look like for the field $K$? Similarly as above we see that any isomorphism $f : K \to K$ we must have $f(\sqrt[3]{a}) = \sqrt[3]{a}\zeta_3^n$ for some integer $n$, and similarly $f(\zeta_3) = \zeta_3$ or $f(\zeta_3) = \zeta_3^2$. We also note that if we know the values $f(\sqrt[3]{a})$ and $f(\zeta_3)$, then we know the values of all $x \in K$. We thus see that there are at most $3 \cdot 2 = 6$ such isomoprhisms for $K$. We denote the set of these maps by $\text{Gal}(K/\mathbb{Q})$, now abbreviated $G$. This set $G$ is actually quite nice: the elements of the set form a group under composition - if $f$ and $g$ are elements of $G$, then the map $h$ defined by $h(x) = f(g(x))$ is also in this set (check!). One can also check that all the other properties of groups are satisfied, so we call $G$ the Galois group of $K$. Note that $f(g(x))$ and $g(f(x))$ are not necessarily the same map! (That is, the group is not necessarily abelian.) The Frobenius density theorem connects the set of primes $p$ for which $P(x) \equiv 0 \pmod{p}$ with the Galois group of $K$. Let $S$ be the set of $f \in \text{Gal}(K/\mathbb{Q})$ which fix at least one root of $P$. Let $q = \frac{|S|}{|G|}$. The Frobenius density theorem says that the density of the primes $p$ for which $P(x) \equiv 0 \pmod{p}$ is solvable exists and equals $q$. The density is just the number of primes $p$ for which the equation is solvable divided over the number of all primes $p$ (we, of course, have to take limits to make this rigorous). In fact, the Frobenius density theorem tells much more - it can be used to answer questions on the irreducible factors of $P$ modulo $p$. The above version, however, suffices for our purposes. How does this apply to our problem? If $q \neq 1$ above, then not all primes $p$ divide some value of $P(x) = x^3 - a$. This would lead to a contradiction. Therefore, we must have $q = 1$, so some root of $x^3 - a$ is always fixed under any isomoprhism of $K$. We quickly prove that this is the case only if $a$ is the cube of some integer. The extension of $\mathbb{Q}(\zeta_3)$ is of degree $2$. The degree is a quantity associated to a field which measures its "size", and equals to the degree of the minimal polynomial of $\zeta_3$. In this case the minimal polynomial is $x^2 + x + 1$. Similarly, the degree of $\mathbb{Q}(\sqrt[3]{a})$ is $3$ for $a$ not a cube of an integer. Now, $K = \mathbb{Q}(\zeta_3, \sqrt[3]{a})$ is a field which has a subfield of size $2$. This means that its degree is divisible by $2$ (this should sound intuitive - one actually needs the multiplicative property of field extensions to make this rigorous). Similarly, the degree is divisible by $3$, so the degree must be divisible by $6$. One easily also sees that the degree is at most $6$, so it is exactly six. What we have proven above is basically just the fact that $\zeta_3$ and $\sqrt[3]{a}$ "do not communicate with each other". This means that for any $x, y$ with $0 \le x \le 2$ and $1 \le y \le 2$ there exists an isomorphism of $K$ which maps $\sqrt[3]{a}$ to $\sqrt[3]{a}\zeta_3^x$ and $\zeta_3$ to $\zeta_3^y$. This means that the Galois group of $K$ consists of all permutations on $3$ elements. It is now easy to see that there exists elements which do not fix any root of $P$, which proves the claim. Remark: the above procedure generalizes for any prime $q$ in place of $3$. The claim does not hold for the the $8$th power residues (an elementary, though non-trivial, exercise!). I have heard that the powers divisible by $8$ are the only ones for which this generalization fails - I believe this to be correct, though I have not thought of the details of the proof. The shortlist problem. In post 5 it was noted that if $ab$ is not a perfect square, then we are done. We generalize this idea for higher powers, which leads to a full solution. Lemma. Let $p$ and $q$ be primes such that $p \equiv 1 \pmod{q}$, $a$ is a $q$th power residue modulo $p$, and $b$ is not a $q$th power residue modulo $p$. Then there exists a positive integer $n$ such that $(a^n - 1)(b^n - 1)$ is not a perfect square. Note that the conditions of the lemma are just a way to guarantee that $ord_p(a) \neq ord_p(b)$. Proof of lemma. The idea is to look at the value $n = \frac{p-1}{q}$. Then $v_p((a^n - 1)(b^n - 1)) = v_p(a^n - 1)$. If this is odd, we are done. Otherwise, look at $n = p\frac{p-1}{q}$, and use LTE as in post 17 to force the oddness of $v_p(a^n - 1)$. We now sketch a proof on how to find such $p$ and $q$ as described by the lemma. We have to divide into two cases. Case 1. $a$ and $b$ are multiplicatively independent, that is, there does not exist integers $n$ and $m$ such that $a^nb^m = 1$, with at least one of $n, m$ nonzero. Case 2. $a$ and $b$ are multiplicatively dependent. Case 1 is the easier one: This case "looks like" $a = 2, b = 3$. The point is that for all large enough $q$ the degree of the extension $\mathbb{Q}(\zeta_3, a^{1/q}, b^{1/q})$ is $(q-1)q^2$. This follows from the results presented here: http://www-users.math.umn.edu/~garrett/m/v/linear_indep_roots.pdf. Now, it is quite direct to prove the existence of a desired isomorphism, which proves that there are infintiely many primes $p$ as desired in the lemma. Case 2 "looks like" $a = 4, b = 8$. The point is to write $a = c^x$, $b = c^y$ with $\gcd(x, y) = 1$. By symmetry, assume that $x > 1$. Take some prime dividing $x$, say $q$. Now, $q$ does not divide $y$, and it can now be proven, as in the solution of our first problem, that for infinitely many $p$ the number $c^y$ is not a perfect $q$th power modulo $p$. Final comment: The Frobenius density theorem is good at handling conditions such that "$a$ should be a $n$th power residue modulo $p$". Its generalization, the Chebotarev density theorem, can also be used to impose congruence conditions for the moduli: "$p$ should be congruent to $x$ modulo $y$". However, for $x \equiv 1 \pmod{y}$ one can just use Frobenius. A recommendation for further reading: https://scholarship.claremont.edu/cmc_theses/993/.

EDIT: as is stated in the below post, a main part of this proof is wrong - maybe it can be corrected but the way to do it isn't obvious to me.

Quote: $2(k+j)\sqrt{ab} + k^2-j^2 = 2 $ That's not true, the correct equality is $2(k-j)\sqrt{ab} + k^2+j^2 = 2. $

I hope my solution correct Fact 1: if $x>k$ and both of them natural numbers then $(a^x+1)^{\frac{k}{x}}>a^k+1$ Fact 2: let $a=b+x$ then we can find large $n$ such that there always exists a positive integer $k$ such that $b^{x+n+k+1}>a^n>b^{x+n+k}$ then, $(a^n+1)^{\frac{x+2n+k}{n}}>(a^{x+n+k}+1)(b^{x+n+k}+1)$ $(a^{x+n+k}+1)(b^{x+n+k}+1)>(a^n+1)^{\frac{x+n+k-1}{n}}\cdot b^c(b^{x+n+k}+1)$ firstly, for every large $n\in N $ is true that $(a^{x+n+k}+1)>(a^n+1)^{\frac{x+n+k-1}{n}} \rightarrow (a^{x+n+k}+1)^n>(a^n+1)^{x+n+k-1}$ Then, $((a^{x+n+k}+1)^n-(a^n+1)^{x+n+k-1})\cdot a<(a^{x+n+1+k}+1)^{n+1}-(a^{n+1}+1)^{x+n+k} $ (We want to prove that left side increasing ) $(((a^{x+n+k}+1)^n-(a^n+1)^{x+n+k-1})\cdot a) \sim (a^{(x+n+k)\cdot n +1} - a^{n(x+n+k-1)+1})<((a^{x+n+1+k}+1)^{n+1}-(a^{n+1}+1)^{x+n+k})\sim (a^{(x+n+k+1)(n+1)} - a^{(n+1)(x+n+k)}) \rightarrow a^{nx+n^2+kn-n+1}(a^n-1)< a^{nx+n^2+nk+x+n+k}(a^{n+1}-1)$ and that means our inequality is correct Thus , left side extremely increasing so $c$ can be large number Next, $(a^n+1)^{\frac{x+2n+k}{n}}>(a^{x+n+k}+1)(b^{x+n+k}+1)>(a^n+1)^{\frac{x+n+k-1}{n}}\cdot b^c(b^{x+n+k}+1) \rightarrow$ $(a^n+1)^{\frac{n+1}{n}}\sim a(a^n+1)>b^{x+n+k+c}+1$ Also, we can calculate the formula that if we icrease $n$ for $j$ than $c$ increasing for $o\geq j$ Additionally $k$ depends on $n$ but it can not increase so extreme as a $c$ so we always can choose such $n$ that exist $c> \frac{x+k+1}{n}+2$ $b^{x+n+k+c}+1>b^{x+n+k+1}\cdot b^{\frac{x+k+1}{n}+1}>(a^n+1)b^{\frac{x+n+k+1}{n}}>(a^n+1)a$(Fact 2) But it is impossible if $x \neq 0$

KOMKZ wrote: I hope my solution correct Fact 1: if $x>k$ and both of them natural numbers then $(a^x+1)^{\frac{k}{x}}>a^k+1$ Fact 2: let $a=b+x$ then we can find large $n$ such that there always exists a positive integer $k$ such that $b^{x+n+k+1}>a^n>b^{x+n+k}$ then, $(a^n+1)^{\frac{x+2n+k}{n}}>(a^{x+n+k}+1)(b^{x+n+k}+1)$ $(a^{x+n+k}+1)(b^{x+n+k}+1)>(a^n+1)^{\frac{x+n+k-1}{n}}\cdot b^c(b^{x+n+k}+1)$ firstly, for every large $n\in N $ is true that $(a^{x+n+k}+1)>(a^n+1)^{\frac{x+n+k-1}{n}} \rightarrow (a^{x+n+k}+1)^n>(a^n+1)^{x+n+k-1}$ Then, $((a^{x+n+k}+1)^n-(a^n+1)^{x+n+k-1})\cdot a<(a^{x+n+1+k}+1)^{n+1}-(a^{n+1}+1)^{x+n+k} $ (We want to prove that left side increasing ) $(((a^{x+n+k}+1)^n-(a^n+1)^{x+n+k-1})\cdot a) \sim (a^{(x+n+k)\cdot n +1} - a^{n(x+n+k-1)+1})<((a^{x+n+1+k}+1)^{n+1}-(a^{n+1}+1)^{x+n+k})\sim (a^{(x+n+k+1)(n+1)} - a^{(n+1)(x+n+k)}) \rightarrow a^{nx+n^2+kn-n+1}(a^n-1)< a^{nx+n^2+nk+x+n+k}(a^{n+1}-1)$ and that means our inequality is correct Thus , left side extremely increasing so $c$ can be large number Next, $(a^n+1)^{\frac{x+2n+k}{n}}>(a^{x+n+k}+1)(b^{x+n+k}+1)>(a^n+1)^{\frac{x+n+k-1}{n}}\cdot b^c(b^{x+n+k}+1) \rightarrow$ $(a^n+1)^{\frac{n+1}{n}}\sim a(a^n+1)>b^{x+n+k+c}+1$ Also, we can calculate the formula that if we icrease $n$ for $j$ than $c$ increasing for $o\geq j$ Additionally $k$ depends on $n$ but it can not increase so extreme as a $c$ so we always can choose such $n$ that exist $c> \frac{x+k+1}{n}+2$ $b^{x+n+k+c}+1>b^{x+n+k+1}\cdot b^{\frac{x+k+1}{n}+1}>(a^n+1)b^{\frac{x+n+k+1}{n}}>(a^n+1)a$(Fact 2) But it is impossible if $x \neq 0$ Absolute cinema