Write it as $x^2+11^2=y^7+2^7=(y+2)(y^6-2y^5+-...+2^6)$. Looking $\mod 4$ gives that $x$ is even and $y$ odd. Now $y+2$ and $(y^6-2y^5+-...+2^6)$ are coprime. But either $y+2$ or $(y^6-2y^5+-...+2^6)$ has an odd number of prime factors $\equiv 3 \mod 4$, thus the product can't be the sum of two squares.

Thank you, ZetaX.

From which IMO shortlist did you find it ?It;s new or old?

IMO 2000 shortlist.

$ n^7+7=a^2$ so $ n^7+2^7=a^2+11^2$ if n is even we have $ a^2 \equiv 7(mod 8)$ which contradicts.so n is odd. prime divisor of $ a^2+11^2$ has the form $ 4k+1$ so $ n+2 \equiv 1(mod 4)$ so $ n \equiv -1(mod 4)$ this means that $ n^7+7 \equiv a^2 \equiv 2(mod 4)$ contradicts.

sumita wrote: $ n^7 + 7 = a^2$ so $ n^7 + 2^7 = a^2 + 11^2$ if n is even we have $ a^2 \equiv 7(mod 8)$ which contradicts.so n is odd. prime divisor of $ a^2 + 11^2$ has the form $ 4k + 1$ so $ n + 2 \equiv 1(mod 4)$ so $ n \equiv - 1(mod 4)$ this means that $ n^7 + 7 \equiv a^2 \equiv 2(mod 4)$ contradicts. How about $ a = 11b$? Here is my proof: Obviously $ n$ has the form $ 4k + 1$. now $ n + 2|a^2 + 121$, therefore $ a^2 + 121$ must have a prime factor of the form $ 4k + 3$, let say $ p$ if $ p$ is not $ 11$, then $ 11^{p - 1} = ( - a^2)^{\frac {p - 1}{2}} = - 1 (mod p)$, contradiction. Therefore $ p = 11$ Then $ 11|n + 2$ and $ a = 11b$. since $ \frac {n^7 + 128}{n + 2} \equiv 9 (mod 11)$, thus $ n = 11^2c - 2$ and $ c|b^2 + 1$ Then we will have $ n \equiv 3 (mod 4)$

sorry for case $ a=11b$ but it is not important because if $ 11|n+2$ then $ 11$ cant divide $ \frac{n^7+2^7}{n+2}$ so we have $ 11^2| n+2$ again we have $ n\equiv -1(m0d 4)$

orl wrote: Prove that for no integer $ n$ is $ n^7 + 7$ a perfect square. If I am not mistaken, a very similar problem appears in Problem Solving Stategies, by Arthur Engel. Here is my soution:

EDIT: Sorry, Mathias_DK . My "proof" is nonsense; the assumptions I made were, indeed, incorrect. EDIT 2: I hope this solution will work (in similar vein to the solution given by Engel in his book. Well, actually, it is the exact same solution, just managed a little to fit the given numbers.):

The QuattoMaster 6000 wrote: Now, if $ 2|(m - \sqrt {7})$, then we would have that $ \frac {m}{2} - \frac {1}{2}\cdot \sqrt {7}$ is an integer in $ \mathbb{Z}\sqrt {7}$, which is not true since it is not in the form of $ a + b\sqrt {7}$ where $ a$ and $ b$ are integers in $ \mathbb {Z}$. Thus, the $ \gcd$ we are calculating is $ \gcd (m - \sqrt {7}, m) = \gcd (m, \sqrt {7})$. Just remember that $ 2 = (3-\sqrt{7})(3+\sqrt{7})$ is not a prime element. Could you explain this step a bit more clearly to me ? Another thing. Are you sure that $ (a,b) = (a-b,b)$ in $ \mathbb{Z}\sqrt{7}$? For example $ (3+\sqrt{7}, 3 - \sqrt{7}) = (3+\sqrt{7}, 6) = (3+\sqrt{7}, 3(3+\sqrt{7})(3-\sqrt{7})) = 3 + \sqrt{7}$ And on the other hand: $ (3+\sqrt{7}, 3 - \sqrt{7}) = (3-\sqrt{7}, 6) = (3-\sqrt{7}, 3(3+\sqrt{7})(3-\sqrt{7})) = 3 - \sqrt{7}$ Is this because $ \mathbb{Z}\sqrt{7}$ doesn't have unique prime factoring? (Fx $ (4-\sqrt{7})(4+\sqrt{7}) = 3^2$

$ \mathbb{Z}[ \sqrt{7} ]$ isn't a UFD. The ring of integers in $ \mathbb{Q}[ \sqrt{7} ]$ is $ \mathbb{Z} \left[ \frac{1 + \sqrt{7}}{2} \right]$, and one can show that this ring, on the other hand, is a UFD. It is possible that that proof is salvageable.

t0rajir0u wrote: $ \mathbb{Z}[ \sqrt {7} ]$ isn't a UFD. The ring of integers in $ \mathbb{Q}[ \sqrt {7} ]$ is $ \mathbb{Z} \left[ \frac {1 + \sqrt {7}}{2} \right]$, and one can show that this ring, on the other hand, is a UFD. It is possible that that proof is salvageable. Sorry, but what is a UFD?

A unique factorization domain is an integral domain with unique factorization into primes.

t0rajir0u wrote: $ \mathbb{Z}[ \sqrt {7} ]$ isn't a UFD. The ring of integers in $ \mathbb{Q}[ \sqrt {7} ]$ is $ \mathbb{Z} \left[ \frac {1 + \sqrt {7}}{2} \right]$, and one can show that this ring, on the other hand, is a UFD. It is possible that that proof is salvageable. No, the original ring is the correct one as $ 7 \not\equiv 1 \mod 4$. thus the ring of integers is $ \mathbb{Z}[ \sqrt {7} ]$. But QuattoMaster's proof fails otherwise: The QuattoMaster 6000 wrote: This means that since they multiply to create a perfect seventh power, they are both seventh powers themselves. This isn't true as you forget about units. Think about $ (-4) \cdot (-9) = 36$: two coprime numbers multiply into a square, but still none of them is a square.

My apologies; I was thinking of $ \mathbb{Z} \left[ \frac{1 + \sqrt{-7}}{2} \right]$.

ZetaX wrote: But QuattoMaster's proof fails otherwise: The QuattoMaster 6000 wrote: This means that since they multiply to create a perfect seventh power, they are both seventh powers themselves. This isn't true as you forget about units. Think about $ ( - 4) \cdot ( - 9) = 36$: two coprime numbers multiply into a square, but still none of them is a square. Sorry, but I don't quite understand how I am forgetting units here. Seventh powers can be negative. Am I misunderstanding what you mean by units? Wait, so does my first solution work through the hole that Mathias_DK described? I am quite new to working in UFD's...

There are more units than $ \pm1$, in fact the units are given by $ \pm (8 - 3 \cdot \sqrt 7)^k$. But the $ \gcd$ properties still hold, so the other mentioned hole isn't one. The mentioned example shows that the given elements differe by a unit factor only: $ 3+\sqrt 7 = (3-\sqrt 7) \cdot (8+3 \sqrt 7)$ and $ 3-\sqrt 7 = (3+\sqrt 7) \cdot (8-3 \sqrt 7)$

sumita wrote: prime divisor of $ a^2 + 11^2$ has the form $ 4k + 1$ Can someone explain this?

Is not completely true...if $ p=4k+3$ divides $ a^2+b^2$ then it divides both $ a$ and $ b$ This is because $ a^2=-b^2 (mod p)$ If $ a=0 (mod p)$ $ b=0(mod p)$ If $ a$ and $ b$ are nonzero you can rewrite it as $ (a/b)^2=-1( mod p)$ But since $ p=4k+3$, $ -1$ is not a quadratic residue. Daniel

My friend told me that modulo 29 works. I haven't checked it though.

FTSOC, assume it is a perfect square. Note that $n$ has to be $1\pmod{4}$. Then: \[n^7+2^7=a^2+11^2.\]We can say $n+2|a^2+121$, so there must be a prime that is $3\pmod{4}$ that divides $a^2+121$. By FCT, this prime must be $11$. Using LTE, we get: \[v_{11}(n+2)\equiv 2.\] However, this means that there are two primes that are $3\pmod{4}$ that divide $n+2$, making it $1\pmod{4}$, which is a contradiction, as desired.

can't you prove this with units digit and then try all the cases

We have $n^7+128=m^2+121,$ take $\pmod 4$ giving $m$ even and $n \equiv 1 \pmod 4.$ However $n+2 \mid n^7+128=m^2+121$ and $n+2 \equiv 3 \pmod 4$ but also $\frac{n^7+128}{n+2} \equiv 3 \pmod{4}$ so if both divide $m^2+121$ then $11$ divides both of them by fermat christmas. However we can check by euclidean algorithm that this implies $11 \mid 64 \cdot 7,$ contradiction.

For the sake of contradiction, assume \[n^7+7 = k^2 \implies n^7+128 = k^2+11^2.\] We note that $k$ must be even, and thus $n \equiv 1 \pmod 4$. However, unless $k$ is divisible by 11, all prime factors $p$ of $k^2+11^2=n^7+128$ must be $1 \pmod 4$, meaning all divisors should also be $1 \pmod 4$. We see $n+2$ is a contradiction. Finally, if $k$ is a multiple of 11, we have $n \equiv 1 \mod 4$ and $121~\Vert~(n+2)(n^6-2n^5+\ldots+64) \equiv 1 \pmod 4$. We quickly get a contradiction if 121 fully divides either of these factors by Fermat's Christmas Theorem, so we must have $11~\Vert~n+2$. By LTE, we get \[v_{11}(k^7+128) = v_{11}(k+2) + v_{11}(7) = 1,\] giving our final contradiction. $\blacksquare$

FTSOC, let $n^7+7=k^2$, this implies that \[n^7+128 = (n+2)(n^6-2n^5+4n^4-8n^3+16n^2-32n+64) = k^2+121.\] Notice that by taking mod $4$, we get that $n \equiv 1 \pmod{4}$. Now, we introduce a useful claim before moving on: Claim: $n+2$ and $n^6-2n^5+4n^4-8n^3+16n^2-32n+64$ are relatively prime. Proof: Begin by noticing that $\gcd(n+2, n^6-2n^5+4n^4-8n^3+16n^2-32n+64) = \gcd(n+2, 448)$, so it suffices to prove that $n \not \equiv 5 \pmod{7}$, since we already know that $n$ is odd. Simply plugging in $n \equiv 5 \pmod{7}$ gives $k^2 \equiv 5 \pmod{7}$, a contradiction as $5$ is a quadratic nonresidue modulo $7$. $\square$ Since $n \equiv 1 \pmod{4}$, we have both $n+2$ and $n^6-2n^5+4n^4-8n^3+16n^2-32n+64$ are $3 \pmod{4}$, so by Fermat's Christmas Theorem, they both must be divisible by $11$, a contradiction to our claim. $\square$

Nt sobad. Note that it is easy to see $n \leq 0$ fail. Now take $n \geq 1$ and assume, $$n^7 + 7 = a^2$$for some square $a$. Now as perfect squares are $0$ and $1$ modulo $4$ we can easily show that $n \equiv 1 \pmod{4}$. Then adding $121$ to both sides we wish to find solutions to $$n^7 + 2^7 = a^2 + 11^2$$Now we factor the left hand side to find $$(n+2)(\dots) = a^2 + 11^2$$and hence there exists some prime congruent to $3$ modulo $4$ dividing the right hand side. Hence there exists some prime $p$ dividing $a^2 + 11^2$. Now by Fermat's Christmas Theorem we must have $p \mid \gcd(a^2,11^2)$. However this implies that $\gcd(a^2, 11^2) > 1$ and namely $p = 11$. Then we have $$n^7 + 2^7 = 11^2(b^2 + 1)$$ Finally note that LTE gives, \begin{align*} \nu_{11}(n^7 + 2^7) = \nu_{11}(n+2) \end{align*} Then as the left hand side has an odd $\nu_{11}$ and the right hand side has an even $\nu_{11}$ we have reached our desired contradiction.

Clearly $n \geq 1$. Taking $\pmod{4}$ we have $n^7 + 3 \equiv k^2\pmod{4}$ so $n \equiv 1\pmod{4}$. I was told to add $121$. $\newline$ \[n^7 + 7 = k^2 \implies n^7 + 128 = k^2 + 11^2\]Since $n$ is odd, $k^2 + 11^2$ is odd, so all prime factors of $k^2 + 11^2$ is $\equiv 1\pmod{4}$ by Fermat's Christmas Theorem. If some prime that divides $k^2 + 11^2$, then it divides $\gcd(k^2, 11^2)$ which is presumably $11^2$, because if otherwise, then $p | \gcd(k^2, 11^2) = 1 \equiv 1\pmod{4}$. $\newline$ We can factor $n^7 + 128$ as \[(n + 2)(n^6 - 2n^5 +\ldots + 64)\]However, $(n + 2) \equiv 3\pmod{4}$ which divides $k^2 + 11^2$ so we can assume that $11 | k^2 + 11^2 \implies 121|k^2$. $\newline$ So, we have \[n^7 + 128 = 11^2(m^2 + 1)\]Taking $\nu_{11}$ results in \[\nu_{11}(n + 2) = \nu_{11}(m^2 + 1) + 2\]\[\nu_{11}(m^2 + 1) = 0\]which is easily verifiable, by taking all possible $m$ modulo $11$. So, \[\nu_{11}(n + 2) = 2\]If $\nu_{11}(n + 2) = 2$, then we have a contradiction since $n + 2 \equiv 3\pmod{4}$ implies there is another $3\pmod{4}$ divisor. But then by Fermat Christmas, $n + 2 | k^2 + 11^2$ implies that the divisor is $11$, so $\nu_{11}(n + 2) > 2$, so we are done.

$n^7+7=k^2 \Rightarrow n^7+128=k^2+121=k^2+11^2$. It's not hard to see, that $n \equiv 1 (\mod{4})$, cause if $2 \mid n$, then $k^2+1 \equiv 0 (\mod{4})$, and if $n \equiv 3 (\mod{4})$, then $k^2 + 1 \equiv 3 (\mod{3})$. Let $p \mid n+2, \Rightarrow p \mid n^7+2^7 = k^2 + 11^2$, as $n+2 \equiv 3 (\mod{4})$, $n+2$ has a prime devisor that also $\equiv 3 (\mod{4})$ (let it be $p$), then: Theorem 1: If $p \mid a^2+b^2$ and $p \equiv 3 \mod{4}$, then $p \mid a,b$ By theorem 1, $11 \mid k, p = 11$. Turns out that $11^2 \mid n^7+2^7 = k^2 + 11^2$, then by LTE, $\nu_{11}(n^7+2^7)=\nu_{11}(n+2)$, then $11^3 \mid n+2$, so $11 \mid \frac{k^2}{11^2} + 1$ which is a contradiction, so there are no such $n \in N$

That's really easy, have $m=n+2$ and then it's easy to see why $$n^7 + 2^7 = m \sum_{1 \le k \le 7} \binom{7}{k} (-1)^{k-1} m^{k-1} (2)^{7-k}$$

Let $n^7+7=a^2$ we can check that $n\equiv 1\pmod 4$; add $121$ to both sides to get $$n^7+2^7=a^2+11^2$$Take $p$ prime such that $p\mid n+2$ then $p\mid n^7+2^7=a^2+11^2$ so by Fermat Christmas $p\mid a,11$ in which case $p=11$ or $p\equiv 1\pmod 4$, if the latter occurs for all primes then $n+2\equiv 1\pmod 4$ which is impossible so $11\mid n+2$ so $$n^7+2^7=11^2(a_1^2+1)$$but $v_{11}(11^2(a_1^2+1))=2$ and by LTE $v_{11}(n^7+2^7)=v_{11}(n+2)$, so $v_{11}(n+2)=2$ but every prime divisor $p$ of $n+2$ is $1\pmod 4$ except for $p=11$ so $n+2\equiv 11^2\equiv 1\pmod 4$ which is impossible

It s easy to see that $n$ is a strictly positive integer, and taking the equation modulo $4$ implies $n \equiv 1 \pmod{4}$. We can now rewrite the equation as $n^7+2^7 = x^2+11^2$. We know that $n+2 \mid n^7+2^7 $, but since $n +2\equiv 3 \pmod{4}$ then $11 \mid n+2$ by Fermat's Christmas theorem. And this leads to a contradiction since $v_{11}(x^2+11^2)=2$ and $v_{11}(2^7+n^7)=v_{11}(n+2)+v_{11}(7) \equiv 1 \pmod{2}$ by LTE. $$\mathbb{Q.E.D.}$$

For the sake of contradiction, assume $n^7 + 7 = a^2$ where $n, a$ are integers. Then $\mod{4}$ easily gives $n \equiv 1 \pmod{4}$. Now we can add $121$ to both sides to obtain $$(n + 2)(n^6 - 2^2n^5 + 2^3n^4 - \dots - 2^5n + 2^6) = n^7 + 2^7 = a^2 + 11^2.$$Note that $n + 2$ is odd. Now Fermat's Christmas Theorem gives that $n + 2$ is either the product of primes that are all $1 \pmod{4}$ or is divisible by $11$. Earlier we derived $n \equiv 1 \pmod{4}$, hence it must be the latter. Now plugging in $n \equiv 9 \pmod{11}$ yields $$9^6 - 2^2 \cdot 9^5 + 2^3 \cdot 9^4 - \dots - 2^5 \cdot 9 + 2^6 \equiv (-2)^6 + 4 \cdot 9^7 + 2^6 \equiv 4 \cdot 9^7 \not \equiv 0 \pmod{11}$$. But it is easy to verify that the above term is neither odd nor $1 \pmod{4}$, hence we arrive at a contradiction.

Suppose $n^7+7=x^2$ for some integral $x.$ Take $n^7+2^7=x^2+11^2.$ $v_2$ the RHS is at most $1$ so $n$ cannot be even (and $x$ is even), and taking mod 4 we have $n \equiv 1 \mod 4$ so there exists a $p \equiv 3 \mod 4$ that divides $n+2.$ By Fermat's Christmas Theorem on the RHS, this can only be $11.$ $v_{11}(n^7+2^7)=v_{11}(n+2)=2$ as it must be $2$ on the RHS. As $n+2$ is not divisible by any other $p \equiv 3 \mod 4$ we find that $n+2 \equiv 1 \mod 4$ which forms a contradiction as we already know $n \equiv 1 \mod 4.$

For the sake of a contradiction, assume that $n^7+7 = k^2$ for a positive integer $k.$ Then the equation is equivalent to $$k^2+11^2=n^7+2^7=(n+2)(n^6-2n^5+2^2n-2^3n^3+2^4n^2-2^5n+2^6).$$Taking mod $4$ yields $n \equiv 1 \pmod 4.$ Therefore, $n+2 \equiv 3 \pmod 4,$ and $n^6-2n^5+2^2n-2^3n^3+2^4n^2-2^6n+2^6 \equiv 3 \pmod 4.$ Hence, there are primes $p, q \equiv 3 \pmod 4$ such that $$p|n+2, q|(n^6-2n^5+2^2n-2^3n^3+2^4n^2-2^6n+2^6) \implies pq|(n^7+2^7).$$But by Fermat's Christmas Theorem this is only possible when $p=q=11.$ Therefore, $n \equiv 9 \pmod {11},$ which means that $$n^6-2n^5+2^2n-2^3n^3+2^4n^2-2^6n+2^6 \equiv 8 \pmod {11},$$a contradiction. QED