an ugly solution First test: $n^2 \equiv 0,1\pmod 4$ We obviate $11\cdots 11, 22\cdots 22, 55\cdots 55, 66\cdots 66, 99\cdots 99$ and $33\cdots 33,44\cdots 44,77\cdots 77,88\cdots 88$ are left. Second test : $n^2 \equiv 1,4,5,6,9\pmod {10}$ Only $44\cdots 44$ remains. if is a square, $A^2 = 44\cdots 44$, thus $4|A^2$ and hence $2|A$, say $A=2B$, thus $B^2 = 11\cdots 11$ which is impossible due to the first test.

yes liyi, with mods it's easy. But since this was for 10th grade max, none had ever heard of mods. there's a simpler solution that any school kid can come up with (if they have brains though )

Let's look at the array vertically. It proceeds as follows: n 2n 3n ... 9n Since n>9, it always stops before n (a perfect square). Now, let's look at the first horizontal line. It's made up entirely of 1s. It's mathematically impossible to multiply two identical integers or two identical fractions that do not equal 1 and end up with such a number. I don't know how to prove this, though.

hun?? I din't get a word of that sorry

111111... can't be a square because it is always 3 mod 4. 222.... can't be a square because it is 2 mod 4 3333... cant be a square ( notice that it is equal to 3*1111...) 4444.... cant be a square because 1111... isnt a square and so on. Is it correct ?

1st row is obviously all divisible by 11, thus they all have an even number of divisors. 2nd row's sum of digits all equal a multiple of three, thus they are all divisible by 3 and have an even number of divisors. 3rd row's alternating digits are all 0, so it is divisible by 11, once again, an even number divisors. Perfect squares have an odd number of divisors.

breez wrote: 1st row is obviously all divisible by 11, thus they all have an even number of divisors. Firstly, if you take a sting of the first row with an odd number of 1's, the number is not divisible by 11. Secondly, how would being divisible by 11 be proof that a number has an even number of divisors? If it were an even power of 11, it would have an odd number of divisors. In your method, it seems like you have to prove that $11^n$ is never a string of all 1's.

Meghrabi wrote: 111111... can't be a square because it is always 3 mod 4. 222.... can't be a square because it is 2 mod 4 3333... cant be a square ( notice that it is equal to 3*1111...) 4444.... cant be a square because 1111... isnt a square and so on. Is it correct ? Just look at that

Peter VDD wrote: yes liyi, with mods it's easy. But since this was for 10th grade max, none had ever heard of mods. there's a simpler solution that any school kid can come up with (if they have brains though ) I'm looking forward to the other solution.

Meghrabi wrote: Meghrabi wrote: 111111... can't be a square because it is always 3 mod 4. 222.... can't be a square because it is 2 mod 4 3333... cant be a square ( notice that it is equal to 3*1111...) 4444.... cant be a square because 1111... isnt a square and so on. Is it correct ? Just look at that Well, maybe there are more people thinking it is not a proof... why does bein 3*11..11 guarantee it's not a square? what with further numbers as 77..77?

Okay then Squares always end on 0,1,4,9,6,5 So we can already cancel 22..,333..,777...,8.... 111... is always 3 mod 4 444..=4*11... so 1111.. has also to be a square and it isn't. 555... is always 3 mod 4 666... is always 2 mod 4 999... is always 3 mod 4 Is it now proved ? If so can you give the other proof ? Thanks

yes, that's better. I do not remember exactly what solution I had in mind when writing that... there are several ways to solve it though. I think it had to do with looking at what number it would be the square of.

Prove, none of the numbers are a perfect power. (except 1st power, ofcourse)

Peter wrote: yes liyi, with mods it's easy. But since this was for 10th grade max, none had ever heard of mods. there's a simpler solution that any school kid can come up with (if they have brains though ) huh? how come mods are teached in high-school only? In Romania the mods are teached in the 5th or 6th grade. Sorry for bad english

Modular arithmetic is often not taught at all in American schools. Mostly because it does not have anything to do with elementary calculus which is what the college track American curriculum is mainly designed to prepare students for.

Let a=11 The list goes a 10a+1 ... a(10)^n+10^n-1 + ... 10^1 +10^0 2a 2(10a+1) ... 2(a(10)^n+10^n-1 + ... 10^1 +10^0) ... ... ... ... 8a 8(10a+1) ... 8(a(10)^n+10^n-1 + ... 10^1 +10^0) 9a 9(10a+1) ... 9(a(10)^n+10^n-1 + ... 10^1 +10^0) For any number in the list to be a perfect square, 0<a(10)^n+10^n-1 + ... 10^1 +10^0 =<9. We know a=11, so we have at least 11(10^0)=11>9 Therefore there are no perfect squares. By the way what do all these mod mean?

A mod is essentially the remainder function. For example, $2=5\mod 3 $ since $\frac53=1\ \mathcal{R}\ 2$

$11=8+3$ $111=108+3$ Not expressible in form $8m+1$ $1111=1108+3$ . And so for next sequence bcoz an even perfect square is of form $4k$ But they are not of form..... And so follows And an odd perfect square is of form $8m+1$ Proof $a^2=(2k+1)^2=4k(k+1)+1=4.2m+1=8m+1$ $a^2=(2k)^2=4k^2=4m$

It remains to prove that none of \[11 \ \ 111 \ \ \ 1111 \ldots\]are perfect squares. By induction, $11$ is not a perfect square, for a number,$a$ in the sequence with $k$ $1$'s that is not a perfect square, the next number $10a+1$ is also not a perfect square since if \[10a+1=j^2 \implies 10a=(j-1)(j+1)\]but $LHS$ is divisible by $2$ as $a$ is odd and $RHS$ is divisible by $4$ as $j$ is odd which is a contradiction.

Peter wrote: Prove that there are no perfect squares in the array below: \[\begin{array}{cccc}11&111&1111&...\\22&222&2222&...\\33&333&3333&...\\44&444&4444&...\\55&555&5555&... \\66&666&6666&...\\77&777&7777&...\\88&888&8888&...\\99&999&9999&...\end{array}\] You can use Modular Arithmetic. For example start with mod 4 and you know all perfect squares are 0,1. mod 32 might be optimal though.

Let $S_n$ denote the infinite set of all positive integers greater than 10 whose only distinct digit is $n$ for all positive integers $n$ from 1 to 9, inclusive. Then note that $S_4$ and $S_9$ contain perfect squares iff $S_1$ contains a perfect square. Similarly, $S_8$ contains a perfect square iff $S_2$ contains a perfect square. Therefore, we just have to prove that none of $S_1, S_2, S_3, S_5, S_6,$ or $S_7$ contain a perfect square. Since all perfect squares end in one of 1, 4, 5, 6, 9, or 0, we now just have to show that none of $S_1, S_5,$ or $S_6,$ contain a perfect square. It's not hard to show that all perfect squares ending in 5 have tens digit 2, so we know that none of the elements of $S_5$ is a perfect square, since all elements of $S_5$ have tens digit 5. Also, we know that each element in $S_6$ has exactly one 2 in its prime factorization, which means that these are also not perfect squares. This means that we just have to prove none of the elements of $S_1$ is a perfect square. We can do this because all of the elements in $S_1$ end in 11, meaning that they are 3 more than a multiple of 4. All squares are either multiples of 4 or are 1 more than a multiple of 4, meaning that we have proven that none of the elements in $S_1$ are squares. Thus, we have proven the original statement. $QED$

Peter wrote: Prove that there are no perfect squares in the array below: \[\begin{array}{cccc}11&111&1111&...\\22&222&2222&...\\33&333&3333&...\\44&444&4444&...\\55&555&5555&... \\66&666&6666&...\\77&777&7777&...\\88&888&8888&...\\99&999&9999&...\end{array}\] firstly we put 2s 3s 7s and 8s out of the game since no square has last number a 2,3,7,8 Then we work with $\frac{A(10^n-1)}{9} = X^2$ which suffices $\frac{A(10^n-1)}{9}$ to be a square there we check the case i) $A=1,4,9$ where $10^n-1$ is a square $10^n= k^2+1$ so for $2\geq n$ no solution, so by $n\geq 3$ $8$ should divide $k^2+1$ which can't happen for any k which is a positive integer ii) we now observe the cases of 5 and 6 For 5 if $5/X^2$ then $5^2/X^2$ and so $5/10^n-1$ impossible since 5 has to divide 1 which is not true For 6 2 should divide $10^n-1$ which is also impossible and so we are done

breez wrote: 1st row is obviously all divisible by 11, thus they all have an even number of divisors. 2nd row's sum of digits all equal a multiple of three, thus they are all divisible by 3 and have an even number of divisors. 3rd row's alternating digits are all 0, so it is divisible by 11, once again, an even number divisors. Perfect squares have an odd number of divisors. no 111 is not divisible by 11

We note that $2,3$ are non-quadratic residues of $4$ as well as $5$. First row can't be a square as it is $3\pmod4$. Second row can't be a square as it is $2\pmod4$. Third row can't be a square as it is $3\pmod5$. Fourth row can't be a square as it is $4$ times a non-perfect square. Fifth row can't be a perfect square as it is $3\pmod4$. Sixth row can't be a perfect square as it is $2\pmod4$. Seventh row can't be a perfect square as it is $2\pmod5$. Eighth row can't be a perfect square as it is $3\pmod5$. Ninth row can't be a perfect square as it is $3\pmod4$.

there just aren't believe me i'm in 6th grade and i came across this problem and i kept grinding i never found a perfect square

mattjunior wrote: there just aren't believe me i'm in 6th grade and i came across this problem and i kept grinding i never found a perfect square why the 6th grade part? also no offense but we really don't need your input on this if you're just going to grind to be honesty modular arithmetic is the absolute go-to solution to this

mathmax12's solution (which he told me to post here): The first row can't be a square because it is $3\pmod{4}$,the second row can't be a square because it is $2\pmod{4}$, the third row can't be a square because it is $3\pmod{5}$, the fourth row can't be a square because it is $4$ times a non perfect square, the fifth row can't be a square because it is $3\pmod{4}$, the sixth row can't be a square because it is $2\pmod{4}$, the seventh row can't be a square because it is $2\pmod{5}$, the eithg row can't be a square because it is $3\pmod{5}$, the ninth row can't be a square because it is $3\pmod{4}$, hence, proved.