If $3n+1$ is a perfect square where $n$ is a positive integer greater than $1$. Then how to prove that $n+1$ would be sum of $3$ perfect squares?
122376
If $n$ is a positive integer greater than 1 such that $3n+1$ is perfect square, then show that $n+1$ is the sum of three perfect squares.  My work: $3n+1=x^2$ $3n+3=x^2+2$ $3(n+1)=x^2+2$ $(n+1)=\dfrac{x^2+2}{3}$  I have no clue what to do next. Please help!
101233
Let $k$ be a natural number. Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.  I tried to prove this by supposing one of them is a square number and by substituting the corresponding $k$ value. But I failed to prove it. If we ignore one term, we can make the remaining terms squares.  For example, $3k+1$ and $4k+1$ are both squares if $k=56$ (they are $13^2$ and $15^2$); $3k+1$ and $6k+1$ are both squares if $k=8$ (they are $5^2$ and $7^2$); $4k+1$ and $6k+1$ are both squares if $k=20$ (they are $9^2$ and $11^2$).
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Assumption: $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}\ge\frac{2}{3}k\sqrt{k}$   Prove true for $n=k+1$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}$$  I'm upto :$$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}k\sqrt{k}+\sqrt{k+1}$$after which I'm stuck.
13594
Can $n!$ be a perfect square when $n$ is an integer greater than $1$?  Clearly, when $n$ is prime, $n!$ is not a perfect square because the exponent of $n$ in $n!$ is $1$.  The same goes when $n-1$ is prime, by considering the exponent of $n-1$. What is the answer for a general value of $n$?  (And is it possible, to prove without Bertrand's postulate. Because Bertrands postulate is quite a strong result.)
113454
I have this really hard problem that I am working on and I just don't seem to get it. The question is: let $n$ be a positive integer; prove that there does not exist any integer $m$ such that $3n^2+3n+7=m^3$.  Please explain in detail otherwise I might not understand. I am also open for any methods that might help. Thank you. Please help.
6444
One observes that  \begin{equation*} 4!+1 =25=5^{2},~5!+1=121=11^{2}  \end{equation*} is a perfect square. Similarly for $n=7$ also we see that $n!+1$ is a perfect square. So one can ask the truth of this question:   Is $n!+1$ a perfect square for infinitely many $n$? If yes, then how to prove.
112406
Show that every positive integer is a sum of one or more numbers of the form $2^r3^s,$ where $r$ and $s$ are nonnegative integers and no summand divides another.   From:   I see they mean that:  $$n-1 = \sum_{k=1}^{x} 2^{r_k} 3^{s_k}$$  By induction hypothesis. Then:  $$n = \sum_{k=1}^{x} 2^{r_k} 3^{s_k} + 1$$  Then how to proceed by $n/2$? How can you say that:  $$\frac{n}{2} = \sum_{k=1}^{x} 2^{r_k} 3^{s_k} $$  Where did the $1$ go? And how can you assume it works for $\frac{n}{2}$ (the hypothesis)?
30493
I'm trying to determine if 1317 and 116 can be written as the sum of three squares?  I have the condition that if it is not of the form $4^{\alpha}(8k+7)$ then it can be written as a sum of three squares, but how do I use this condition to work it out?  Thank you
82941
I'd appreciate your help showing that if $p\equiv3\,\left(\mbox{mod 4}\right)$   then p can't be written as a sum of two squares.  Thanks!
96870
Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is  $\frac{n(n+1)(2n+1)}{6}$.  Can someone pls help and provide a solution for this and if possible explain the question
40058
If $1/a  +  1/b  =  1 /c$   where $a, b, c$ are positive integers with no common factor, $(a + b)$ is the square of an integer.  I found this question in RMO 1992 paper !  Can anyone help me to prove this ?  May be it's too easy and simple . But , I'm just novice in number theory !
48695
I'm having some trouble figuring out where to start for this proof.     Prove for all positive integers $n$,   $$\sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}$$   Any help will be greatly appreciated.
60101
Let $n \in \mathbb N^+$. Show that if $x_1, x_2, ... , x_n$ are $n$ real numbers such that $-1 \le x_i \le 0$ for each $1 \le i \le n$, then $$(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$$  I am using a proof by induction, but I am unable to complete it. I would appreciate any tips on how I could finish the proof.   Incomplete Proof Let $P(n)$ be the proposition that $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + \sum_{i = 1} ^ n x_i$. When $n = 1$, LHS = RHS = $1 + x_i$. Clearly $1 + x_i \ge 1 + x_i$ and so $P(1)$ is true. Assume that $P(k)$ is true, i.e. $(1 + x_1)(1 + x_2)...(1 + x_k) \ge 1 + \sum_{i=1}^k x_i$. Case 1 Suppose $x_{k + 1}  = -1$. Then $1 + x_{k + 1} = 0$, hence LHS = $(1 + x_1)(1 + x_2)...(1 + x_k)(1 + x_{k+1}) = 0$. RHS = $1 + x_1 + x_2 + ... + x_k - 1 = x_1 + x_2 + ... + x_k$ which is always nonpositive since each $x_i$ is nonpositive. Hence RHS $\le 0$ and so LHS $\ge$ RHS. Case 2 Suppose $x_{k+1} = 0$. Then $1 + x_{k+1} = 1$, hence multiplying it to LHS does not change anything. Also, adding $x_{k+1}$ to RHS does not change anything. Therefore, LHS $\ge$ RHS still holds. Case 3 Suppose $-1 &lt; x_{k+1} &lt; 0$. Then $0 &lt; 1 + x_{k+1} &lt; 1$. Adding $x_{k+1}$ to RHS makes the RHS smaller since $x_{k+1}$ is negative. If LHS $= 0$, then multiplying $1 + x_{k+1}$ to LHS does not change anything, and so LHS $\ge$ RHS holds. If LHS > $0$, then multiplying $1 + x_{k+1}$ to LHS makes the LHS smaller since $0 &lt; 1 + x_{k+1} &lt; 1$. In this case LHS $\ge$ RHS does not necessarily hold? ...
37431
Find the sum of all positive integers $n$ for which $|2^n + 5^n– 65|$ is a perfect square.  It is easy to check for $n&lt;3$ only $n=2$ works. Looking at $n≥3$, $2^n + 5^n– 65= a^2$ For odd $n$, LHS has unit digit $2$ or $8$ but RHS has unit digit $0,4$ or $6$. Hence $n$ cannot be odd. For even $n$ we can rewrite our expression as follows ($n=2k$):  $$(a – 2^k)(a+2^k)=5(5^{2k–1} – 13)$$ But I wasn't able to finish off after this. Both LHS and RHS have the same unit digit ($0$). Through trial and error I found that $n=4$ works, but $n=6,8$ don't. Since the questions asks for the sum as is common in Olympiad number theory it's highly unlikely that any other larger values satisfy. I also tried some alternative ways like looking at the equation modulo $4$ but that didn't help.
16879
There are given integers $a, b, c$ satysfaying $a+b+c=0$. Show that $32(a^4+b^4+c^4)$ is a perfect square.  EDIT: I found solution by symmetric polynomials, which is posted below.
21066
We have the sum of squares of $n$ consecutive positive integers: $$S=(a+1)^2+(a+2)^2+ ... +(a+n)^2$$ Problem was to find the smallest $n$ such, that $S=b^2$ will be square of some positive integer. I found an example for $n=11$. Now, I'm trying to prove, that if $2&lt;n&lt;11$ there is't any solution. So I need some help with this: If $n,a,b \in \mathbb{N}$ and $n&lt;11$ prove that equation  $$ n\cdot a^2 + n(n+1)\cdot a + \frac{1}{6}n(n+1)(2n+1)=b^2 $$  can't be solved. Or if I'm wrong, find counterexample.  The only idea I have is: to consider the remains $\operatorname{Mod}[b^2,n]$ and $\operatorname{Mod}[\frac{1}{6}n(n+1)(2n+1),n]$. For each $2&lt;n&lt;11$, but it is very long.
27225
Can a number with $100$ $0$'s $100$ $1$'s and $100$ $2$'s ever be a perfect square? I have no idea where to even start. Some hints would be nice. Thanks in advance :)