So we want $7 \nmid ab(a+b)$ and $7^7 | (a+b)^7-a^7-b^7 = 7ab(a+b)(a^2+ab+b^2)^2$, so we want $7^3 | a^2+ab+b^2$. Now take e.g. $a=2,b=1$ and get $7|a^2+ab+b^2$. Now by some standard methods like Hensels Lemma (used to the polynomial $x^2+x+1$, so $b$ seen as constant from now) we get also some $\overline{a}$ with $7^3 | \overline{a}^2+\overline{a}b+b^2$ and $\overline{a} \equiv a \equiv 2 \mod 7$, so $7\nmid \overline{a}b(\overline{a}+b)$ and we are done. (in this case it gives $\overline{a}=325$)

I found in without Hensels Lemma (which I don't know ), just with few computations that putting $b=1$, $a\equiv_7 2$ and $a^3\equiv_{7^3}1$ gives $a=2^{\cdot 7^2}$ which works... so $a\equiv_{7^3}324$... so $(a,b)=(324,1)$ is a solution

can someone fully explain the solution to this problem? i dont fully understand ZetaX's solution. a=325 is not a solution the smallest solution would be 18,1 is there a way to find all the solutions?

$325 \to 324$ Hensel guarantates us to find all solutions, simply by it's way of working. You can take a look at it on the web.

ZetaX wrote: $325 \to 324$ Hensel guarantates us to find all solutions, simply by it's way of working. You can take a look at it on the web. sorry, i dont understand, could you explain better?!? isnt $\overline{a}$ also a solution?

Kalva wrote: We find that $(a+b)^{7}-a^{7}-b^{7}= 7ab(a+b)(a^{2}+ab+b^{2})^{2}$. So we need find a, b such that $a^{2}+ab+b^{2}$ is divisible by $7^{3}$. At this point I found $a = 18$, $b = 1$ by trial and error. A more systematic argument turns on noticing that $a^{2}+ab+b^{2}= \frac{a^{3}-b^{3}}{a-b}$, so we are looking for $a$, $b$ with $a^{3}\equiv b^{3}\pmod{73}$. We now remember that $a^{\varphi(m)}\equiv 1 \pmod m$. But $\varphi(73) = 2 \cdot 3 \cdot 49$, so $a^{3}\equiv 1 \pmod{343}$ if $a = n^{98}$. We find $2^{98}\equiv 18 \pmod{343}$, which gives the solution 18, 1. This approach does not give a flood of solutions. $n^{98}= 0, 1, 18,$ or $324$. So the only solutions we get are $(1, 18); (18, 324); (1, 324)$. [Moderator: Added Source]

After heavily editing of the previous post, I've time to answer now I think the explaination at http://en.wikipedia.org/wiki/Hensel%27s_lemma is good enough, tell me if you have specific problems.

chien than wrote: We find that $(a+b)^{7}-a^{7}-b^{7}= 7ab(a+b)(a^{2}+ab+b^{2})^{2}$. So we need find a, b such that $a^{2}+ab+b^{2}$ is divisible by $7^{3}$. At this point I found $a = 18$, $b = 1$ by trial and error. A more systematic argument turns on noticing that $a^{2}+ab+b^{2}= \frac{a^{3}-b^{3}}{a-b}$, so we are looking for $a$, $b$ with $a^{3}\equiv b^{3}\pmod{73}$. We now remember that $a^{\varphi(m)}\equiv 1 \pmod m$. But $\varphi(73) = 2 \cdot 3 \cdot 49$, so $a^{3}\equiv 1 \pmod{343}$ if $a = n^{98}$. We find $2^{98}\equiv 18 \pmod{343}$, which gives the solution 18, 1. This approach does not give a flood of solutions. $n^{98}= 0, 1, 18,$ or $324$. So the only solutions we get are $(1, 18); (18, 324); (1, 324)$. Please dont copy and paste other solutions!! this is a copy of the Kalva's solution: http://www.kalva.demon.co.uk/imo/isoln/isoln842.html at least post the source!

That's poor, indeed. And it explains the mass of really stupid and foreseeable $\LaTeX$ mistakes I had to remove.

ZetaX wrote: After heavily editing of the previous post, I've time to answer now I think the explaination at http://en.wikipedia.org/wiki/Hensel%27s_lemma is good enough, tell me if you have specific problems. sorry, im bad at NT , this is as far as i get it: ok, so let $f(a)=a^{2}+a+1$ (b=1) then $f(2)\equiv 0 \pmod{7}$ so by Hensels lemma there exists an integer $0\le \overline{a}\le 6$ s.t. $f(2+7\overline{a})\equiv 0 \pmod{7^{2}}$ or $(2+7\overline{a})^{2}+2+7\overline{a}+1\equiv 0\pmod{7^{2}}$ now what? i dont see how you get $\overline{a}=325$?

orl wrote: Find one pair of positive integers $ a,b$ such that $ ab(a + b)$ is not divisible by $ 7$, but $ (a + b)^7 - a^7 - b^7$ is divisible by $ 7^7$. By writing everything in terms of $ a+b$ and $ ab$ we easily get $ 7^3 \mid a^2+ab+b^2$. Setting $ z = ab^{-1}$, where $ b^{-1}$ is the inverse of $ b$ $ \bmod 7$. We get $ 7^3 \mid z^2+z+1 \mid 4z^2+4z+4 = (2z+1)^2+3$. Then first $ 7 \mid (2z+1)^2+3$, gives a solution $ z \equiv 2 \bmod 7$. Then substitute $ z = 7k+2$, and works $ \bmod 7^2$, and repeat this 'till you get a solution. I found $ z = 37$. Then $ (a,b) = (37,1)$ is a solution

How to get all solutions to this problem?

@3above, this isn't exactly an NT NT problem, at least the way I did it... it's more like expand everything, group together and cancel terms till you get $343 \mid a^2 +ab+b^2$, and since you just need to find $1$ solution, arbitrarily put $b=1$ and bash it out to hit $18$ as your first solution- actually, it's not even a proper bash , as it's pretty simple to estimate first solution should be kind of around of $18$ (as $324$ is pretty close to $343$)... btw the stats for this year is pretty... well, gold was a $40$

At the IMO I did something like this to find a pair $(a,b)$ such that $7^3\big|a^2+ab+b^2$. First realized that either $a\equiv 2b\pmod7$ or $b\equiv 2a\pmod7$; I chose the first one. Then I took the square of $a-2b$: $$ 0 \equiv (a-2b)^2 = a^2+ab+b^2 -b(5a-3b) \pmod{49} $$so we need $49\big|5a-3b$. Took squares again: $$ 0 \equiv (5a-3b)^2 = 25(a^2+ab+b^2)-b(55a-34b) \pmod{7^4} $$... and then I was the stranger who submitted $(34,55)$ instead $(1,18)$. It could be done in one step, by taking cubes: $$ 0 \equiv (a-2b)^3 = (a-7b)(a^2+ab+b^2) + b^2(18a-b) \pmod{7^3} $$so, under the assumption $a\equiv 2b\pmod 7$, we can see that $7^3\big|a^2+ab+b^2$ is equivalent with $18a\equiv b\pmod{7^3}$.

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Note that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$, so $7^7\mid (a+b)^7-a^7-b^7$ is equivalent to $7^3\mid a^2+ab+b^2$. $a=18$ and $b=1$ satisfies this because $18^2+18\cdot 1 + 1^2 = 343=7^3$ and $7\nmid 18\cdot 1 \cdot (18+1)$. So one solution is $\boxed{(18,1)}$.

You can not always assume a Coefficient be 1 (0, -1, etc.) and find it works. I solve "a^2+a*b+b^2" by the method of exhaustion (Code). 1 18 2 36 3 54 4 72 5 90 6 108 8 144 9 162 10 153 11 134 12 115 13 96

Pick $b=1$, so $(a+1)^7 - a^7 - 1 = 7a(a^5 + 3a^4 + 5a^3 + 5a^2 + 3a + 1) = 7a(a+1)(a^4 + 2a^3 + 3a^2 + 2a + 1) = 7a(a+1)(a^2+a+1)^2$. Hence it suffices to ensure $a^2+a+1 \equiv 0 \pmod {7^3}$. Try $a^2+a+1 = 7^3$ - it works for $a=18$.