This is just like ARO 2005, Grade 10, Problem 3 (one of my favorite problems). But in that problem, the exponent of $y$ is $15$, not $14$.

Take a prime $p\equiv 3\pmod{4}$, and consider the Diophantine equation $2x^2 - y^{2p} = 1$, so $2x^2 = (y^2+1)\left ((y^2)^{p-1} - (y^2)^{p-2} + \cdots - y^2 + 1\right )$. The greatest common divisor of the two factors of the RHS must divide $p$, but since $y^2+1 \not\equiv 0 \pmod{p}$, it follows it must be $1$. The case $y^2+1 = a^2$ leads to $y=0$, and no solution. So we must have $y^2+1 = 2a^2$ (a Pell equation with infinitely many solutions), and, denoting $z=y^2 > 0$, also $z^{p-1} - z^{p-2} + \cdots - z + 1 = b^2$, with $x=ab$. Let us quickly solve it for $p=3$, in order to show the way to proceed (placing the expression between consecutive perfect squares). Then we must have $z^2 - z + 1 = b^2$, or $(2z-1)^2 + 3 = (2b)^2$. But $(2z-1)^2 < (2z-1)^2 + 3 < (2z)^2$ for $z > 1$, and so the only solution is $z=1$, $b=1$, when we have $\boxed{x,y \in \{-1,1\}}$. For the case $p=7$, the equation $z^6 - z^5 + z^4 - z^3 + z^2 - z + 1 = b^2$ writes as $(16b)^2 = (16z^3 - 8z^2 + 6z - 5)^2 + 140z^2 - 196z + 231 > (16z^3 - 8z^2 + 6z - 5)^2$. On the other hand, $(16b)^2 = (16z^3 - 8z^2 + 6z - 4)^2 - (32z^3 - 156z^2 + 208z - 240) < (16z^3 - 8z^2 + 6z - 4)^2$ for $z\geq 4$, so it only remains to check the values $z\in \{1,2,3\}$. But $z=y^2$ is a square, so the only possibility is $z=1$, whence $\boxed{x,y \in \{-1,1\}}$ as only solution(s). Note. Finding $16z^3 - 8z^2 + 6z - 5$ can be done by coefficients' identification, but also via the quicker method of using the generalized Newton formula for $\sqrt{z^6 - z^5 + z^4 - z^3 + z^2 - z + 1} = z^3\sqrt{1 - \left (\dfrac {1} {z} - \dfrac {1} {z^2} + \dfrac {1} {z^3} - \dfrac {1} {z^4} + \dfrac {1} {z^5} - \dfrac {1} {z^6}\right )}$, stating $\sqrt{1 - x} = 1 - \dfrac {1} {2} x - \dfrac {1} {8} x^2 + \cdots$ for $|x| < 1$.

Is there any simpler solution?

Seeing this is Problem 6 at Zhautykov, the solution presented is simple enough - in fact it's so straightforward that makes the problem a little unsuitable for its dominating placement in the test

This is a new issue problem from Math Reflections. You should not solve this, but as it has been solved, I am giving my procedure. First show that if a prime $p$ divides $x$ ($x>1$ and assume $x$ positive) then it is of the form $4k+1$. Also $2$ can't divide $x$. Now write the equation as $2(x^2-1)=y^{14}-1$. I now just write $x=4k+1$ as all the divisors are of the form $4k+1$. Now just consider power of $2$ in the equation's left and right sides to arrive at a contradiction. Now the only case is left is $x=1$, which gives $y=\pm 1$. And as if $x$ is a solution, then also $-x$ is, so all possible solutions are combinations of $1$ and $-1$.

How you received a contradiction?

Amir Hossein wrote: Find all integer solutions of the equation the equation $2x^2-y^{14}=1$. Incomplete solution , Anyone can follow these step and solve problem please

At first consider the equation$2x^2-z^2-1=0$ .for $(x_{0},y_{0})=(1,1)$this equality holds.We want to find rational solutions of this equation.consider the curve $2x^2-z^2-1=0$. if $(x,y)$ is a rational point of this curve then $y-y_{0}=m(x-x_{0})$( $(x_{0},y_{0})$ is a obvious point of curve). in this case $(x_{0},y_{0})=(1,1)$ so $y=m(x-1)+1$ ($m$ is a integer). Now return to the original equation. $2x^2-(m(x-1)+1)^2-1=0 \Rightarrow (x-1)(2x-m^2x+m^2-2m)=0$ .Consequently, all rational solutions of the equation are in the form $x=m^2-2m+2/m^2-2$ and $x$ is integer so $m^2-2\mid m^2-2m+2$ .and so $x\in \left \{ 1,-1,-5,+5 \right \}$. we know that $z=y^7$ so $ x\in \left \{ 1,-1 \right \}$. How can we participate in IZO 2013???the problems of this test is reality good .

I think we can use Pelya's equation: $a^2-2x^2=-1$ where $a=y^7$

mavropnevma wrote: Take a prime $p\equiv 3\pmod{4}$, and consider the Diophantine equation $2x^2 - y^{2p} = 1$, so $2x^2 = (y^2+1)\left ((y^2)^{p-1} - (y^2)^{p-2} + \cdots - y^2 + 1\right )$. The greatest common divisor of the two factors of the RHS must divide $p$ ... hello, im new here, please can you explain to me why '' The greatest common divisor of the two factors of the RHS must divide $p$ " ?

Those two factors of the RHS are $y^2+1$ and $(y^2)^{p-1} - (y^2)^{p-2} + \cdots - y^2 + 1$. Say their $\gcd$ is $d$, so $y^2+1 \equiv 0 \pmod{q}$, or $q^2 \equiv -1 \pmod{d}$. But then $0\equiv (y^2)^{p-1} - (y^2)^{p-2} + \cdots - y^2 + 1\equiv$ $ (-1)^{p-1} - (-1)^{p-2} + \cdots - (-1) + 1 = p\pmod{d}$, so we need $d\mid p$. Please, try your brain and hand before, just a little ... and only quote the meaningful part for your question. I removed the unnecessary part.

shekast-istadegi wrote: $(x_{0},y_{0})=(1,1)$ so $y=m(x-1)+1$ ($m$ is a integer). $m$ is a rational number... not an integer.

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IstekOlympiadTeam wrote: if $x\geq2$ $2x^2\equiv0mod4$ Not really.

IstekOlympiadTeam wrote: i think it is so easy $2x^2-y^{14}=1$ if $x\geq2$ $2x^2\equiv0mod4$ so $-y^{14}\equiv1mod4$ which is imposible If $x\geq 2$ and $x$ is odd, then $2x^2 \equiv 2 \pmod{4}$. What you claim only holds for $x$ even, when obviously there is no solution. Did you really think such an easy (as you think it was) question was actually asked in a reputable international competition?

Solution __________________________________________________________________________ Lemma 1. If $a>1$ is an integer then $a^6-a^5+a^4-a^3+a^2-a+1$ is not the square of integer Proof. If $a^6-a^5+a^4-a^3+a^2-a+1$ is an exact square then \[256(a+1)^2(a^6-a^5+a^4-a^3+a^2-a+1)=256(a^8+a^7+a+1)\]is also an exact square. That is impossible.In fact , if $a\geq 3$ then \[(16a^4+8a^3-2a^2+a-1)^2<256(a^8+a^7+a+1)<(16a^4+8a^3-2a^2+a)^2\]if $a=2$ then $a^6-a^5+a^4-a^3+a^2-a+1=43$ Lemma 1. is proved. __________________________________________________________________________ Lemma 2. If $a$ is an integer then $(a+1,a^6-a^5+a^4-a^3+a^2-a+1)$ equals to $1$ or $7$ Proof. The difference $a^6-a^5+a^4-a^3+a^2-a+1-7=(a^6-1)-(a^5+1)+(a^4-1)-(a^3+1)+(a^2-1)-(a+1)$ is divisible by $a+1$. Hence , if $(a+1,a^6-a^5+a^4-a^3+a^2-a+1)=d$ , then $d|7$ Lemma 2 is proved. __________________________________________________________________________ We have \[2x^2=(y^2+1)(y^{12}-y^{10}+y^8-y^6+y^4-y^2+1)\]If $y$ is integer then $y^2+1$ is not divisible by $7$ . By Lemma 2 \[(a+1,a^6-a^5+a^4-a^3+a^2-a+1)=1.\]Hence one of this two numbers is an exact square , the other is an exact square multiplied by 2 . But $y^{12}-y^{10}+y^8-y^6+y^4-y^2+1$ is odd. Therefore, $y^12-y^10+y^8-y^6+y^4-y^2+1=v^2$. According to Lemma 1, $y^2\le 1.$ $y=0$ does not satisfies the condition. Hence , $y\pm 1$ and $x\pm 1$

Notice that we can rewrite the equation in the form : $$2x^2-2y^7=y^{14}-2y^7+1$$Or $2(x^2-y^7)=(y^7-1)^2$ so if we let $x^2-y^7=a\ge0$ ($a$ is positive since it is equal to a perfect square) and $y^7-1=b$ then we get the equation $2a=b^2$. By symmetry we may assume that $x,y$ are positive and hence $a,b$ are nonnegative integers. It is clear that the equation $2a=b^2$ has not any solution exempt $a=b=0$ by infinite descent. So $b=y^7-1=0 \implies y=1 \implies x=1$. Thus all solutions are : $$(x,y)=(1,1),(1,-1),(-1,1),(-1,-1)$$ Please correct me if there is any wrong

I think that my solution is the simplest solution !! Is not it ???

Euleralfeel wrote: I think that my solution is the simplest solution !! Is not it ??? Equation $2a=b^{2}$ has infinity many solutions in naturals.

It's easy to solve this provlem by this lemma: if a,b,c,d are natural numbers satisfying to ab=cd then we can write a=mn, b=pq, c=mp, d=nq and (p,n)=1 where m,n,p,q are natural

Rearranging the initial equation, we get: \[2x^2 = 1+y^{14}\] This implies \[y^{14}+1 = 0 mod 2\]Therefore, \[y^{14}\]is odd and therefore is also the square of an odd number \[\Rightarrow y^{14} = 8m+1\]for some integer m. \[2x^{2} = 8m+2 \Rightarrow x^{2} = 4m+1\]We know if x^2 is a square of an odd number, \[x^{2} = 8k+1\]for some integer k Therefore \[m = 2k\] Now we have \[y^{14} = 16k+1\]and \[x^2 = 8k+1\] Now let \[x^2 = a\]and \[y^{14} = b\] We get the diophantine equation with trivial solution (a,b) = (1,1) and the general solution \[ a = 1 + t\]and \[b = 1 - 2t\]for an integer t. Now we put \[x^2 = a\]and \[y^{14} = b\] We get \[ t = 8k\]and \[t = -8k\]which implies \[k=0\] Therefore there are 4 solutions: \[(\pm1,\pm1)\]