Very interesting problem. How many succeded in solving it Myth?

How could I know that? BTW, you can send your solution here.

Well I didn't solve it to the end but .... The number of perfect squares that have their 17th digit equal to $8$ respectively $7$ are : $M = \sum\([\sqrt{abc9*10^{17}-1}]-[\sqrt{abc8*10^{16}}+1])$ and $N = \sum\([\sqrt{abc8*10^{17}-1}]-[\sqrt{abc7*10^{16}}+1])$ where $abc$ are all numbers with no digit, $1,2,3$ digits. And I am pretty sure that $M > N$

Your idea is absolutely correct! But... It seems you did at least 3 mistakes!

I think I have finished solving it : We will prove that for each $abc$ a small part from $M$ is greater than its relative from $N$ or I will try to prove that : $\sqrt{10A+10^{17}}+\sqrt{A-10^{16}}-\sqrt{A}-\sqrt{10A} \geq 2$ where $A=abc8*10^{16}$ or if we divide all by $\sqrt{10^{16}}$ we have to prove that : $\sqrt{10B+10}+\sqrt{B-1}-\sqrt{B}-\sqrt{10B} \geq \frac{2}{10^{8}}$ where $B=abc8$. I will prove even more : $\sqrt{10B+10}+\sqrt{B-1}-\sqrt{B}-\sqrt{10B} > 0$ or $\sqrt{10B+10}+\sqrt{B-1} > \sqrt{B}+\sqrt{10B}$ or $9 +2\sqrt{10B^2-10} > 2\sqrt{10B^2}$ or $41 + 40B^2+36\sqrt{10B^2-10} > 40B^2$ which is clearly true so $M > N$

Mistakes, mistakes...

Where can I get the questions and full solutions to previous Russian MO competitions. The questions are very very interesting. Thank you.

You can look for Titu Andreescu book World Math Olympiads (I am not sure in this title).

Thank you. I'll look up title. I'm sure I can find it even if information is a little off. Thanks!

I found it. Will probably wait until 2004 edition is out since 2003 is the latest so far. Thanks for the recommendation.

$\forall n<10^{10}, (n+1)^2-n^2<2.10^{10}+1$ If it exists $a,b,c, N$ such that $abc7\times 10^{16}+N$ is a square, we just have to chose in the $\Big[\frac{abc9\times 10^{16} -abc8\times 10^{16}}{2.10^{10}+1}\Big]$ (at least) squares in $[abc8.10^{16}; abc9.10^{16}-1]$ So informations given in the problem are not enough to have the sign of $M-N$ $280000000^2 < 290000000^2$ $78400000000000000 < 84100000000000000$ $290000000^2 < 420000000^2$ $84100000000000000 < 176400000000000000$

Oups I had completly misunderstood the problem, now it's ok. I'm sorry

Let $N_{abc}$ and $M_{abc}$ the number of perfect squares of the form $abc7\times 10^{16} + Q$ and $abc8\times 10^{16} + R$ with $Q, R<10^{16}$ So $N_{abc}$ is the number of perfect squares in $[abc7\times 10^{16};abc8\times 10^{16}-1]$, in this interval the difference D between 2 consecutives squares verify $2.abc7\times 10^{16}+1\leq D\leq 2.abc8\times 10^{16}-1$ Consequently: $\frac{abc8\times 10^{16}-abc7\times 10^{16}}{2.abc7\times 10^{16}+1}\geq N_{abc}\geq \frac{abc8\times 10^{16}-abc7\times 10^{16}}{2.abc8\times 10^{16}-1}$ $\Leftrightarrow \frac{10^{16}}{2.abc7\times 10^{16}+1}\geq N_{abc}\geq \frac{10^{16}}{2.abc8\times 10^{16}-1}$ using same reasonment for $M_{abc}$ we have: $\frac{10^{16}}{2.abc8\times 10^{16}+1}\geq M_{abc}\geq \frac{10^{16}}{2.abc9\times 10^{16}-1}$ Thus $N_{abc}>M_{abc}$ so $N>M$

Inequality $\frac{abc8\times 10^{16}-abc7\times 10^{16}}{2.abc7\times 10^{16}+1}\geq N_{abc}\geq \frac{abc8\times 10^{16}-abc7\times 10^{16}}{2.abc8\times 10^{16}-1}$ is not very good, I think we need to use integer part and add or remove $1$ but the intervall is enough big to keep inequality true

I am not sure I understood it

Using intuition my reasonment is that between $abc7\times 10^{16}$ and $abc8\times 10^{16} - 1$ there is more perfect squares than between $abc8\times 10^{16}$ and $abc9\times 10^{16} - 1$ since the difference between to consecutives squares is less between $abc7\times 10^{16}$ and $abc8\times 10^{16} - 1$ than between $abc8\times 10^{16}$ and $abc9\times 10^{16} - 1$

You know what? I'll let Myth first post if it is correct then I'll rack my brain on it. It looks correct but there is so much too it. Sorry.