The question. Can every $n\in \mathbb N$ can be written: $$n=a^2\pm b^2\pm c^2$$ where $\pm$ are signs of your choice?  We know with  that every integer can be written as the sum of four squares. Plus, with have  stated that an integer can not be written as the sum of three squares if, and only if, it is of the form: $$4^k(8n+7).$$ So we just have to prove (or disprove) it for every number of this form. I have checked it until $55$, and it seems to work so far. So the number we have to check are . For instance: $$31=6^2-2^2-1^2$$ and $$39=6^2+2^2-1^2.$$ The issue here is that $a$, $b$ and $c$ can be arbitrarily large. For instance: $$183=14542^2-14541^2-170^2.$$ So I don't really know how to prove or disprove this result, and I think it could go either way.
123649
Let $Q$ be the ternary quadratic form $Q(x,y,z)=x^2+y^2-z^2$. Since $Q(0,p+1,p)=2p+1$ and $Q(1,p+1,p)=2p+3$, we see that for every integer $k$, the equation $E_k:Q(x,y,z)=k$ always has a solution. Is it known for which integers $k$ there are infinitely many solutions to $E_k$ ?
24340
Show that there are infinitely many positive integers $N$ that cannot be written in the form $a^n+b^n+c^n$ where $a,b,c$ are positive integers and $n \geq 2$.   Note that in the question above, $n$ can vary and is not fixed.  I thought about using a modular arithmetic argument by finding a set of values which are not expressible when the exponent is $3n,3n+2,$ or $3n+4$ and then use the Chinese Remainder Theorem.  Note that for all expressions of the form $$a^{3n}+b^{3n}+c^{3n},$$ the numbers which are $4,5,6$ modulo $9$ are not expressible in these forms since each of the terms is $-1,0,1 \pmod{9}$. For expressions of the form $$a^{3n+2}+b^{3n+2}+c^{3n+2} = a^{3n} \cdot a^2+b^{3n} \cdot b^2+c^{3n} \cdot c^2,$$ note that $x^2 \equiv -1,0,1,4 \pmod{9}$ and that $x^3 \equiv -1,0,1 \pmod{9}$. But I couldn't find a set of numbers modulo $9$ which weren't expressible in this form.   For expressions of the form $$a^{3n+4}+b^{3n+4}+c^{3n+4},$$ note that $x^4 \equiv 0,1,3,9 \pmod{13}$ and $x^{3n} \equiv -1,0,1,8 \pmod{13}$. Therefore, $x^{3n+4} \equiv 0,1,3,7,8,9,10,11,12$, but I couldn't find a set of numbers modulo $13$ which weren't expressible in this form.  Can we use a modular arithmetic argument similar to this to solve the cases for $3n+2$ and $3n+4$ one or do we need to use a higher exponent such as $6n$ instead of $3n$?
44723
Let the sum of all integers $n$ such that $(2n^2+9/n+3)$ is an integer be $A$, what is $|A|$?  I tried expressing it to:  $k$ is an integer, $(n+3)k + 0 = 2n^2 + 9$, $n(2n-k) + 3(3-k) = 0$, from which I couldn't move on,  I also tried $(n+3)k + 0 = 2n^2 + 9$, $k = 2n$, $2n^2 - 2n^2  + 9 - 6n$. $9-6n/n+3$ is an integer, $9 - 6n = 3(3-n)$ though now I also do not know how to move on.
97230
If $n&gt;1$ is an integer not of the form $6k+3$, Prove that $n^2 + 2^n$ is composite.  Any idea of how to think about this problem? I have been thinking about it a lot and yet I was not able to come up with anything.
58624
I am trying to prove old test exercises and this one is resisting me.  I know you can believe that I meant $a^ 2 + b ^ 2$ is divisible by 4, but no, in the test it says that 4 is divisible by $a ^ 2+ b^2$.  Suppose that $ a,b \in \mathbb{Z}$  If 4 is divisible by $ a^2 + b^2$, then $a$ and $b$ are not odd.  I started like this:  Suppose that a and b are even, then $a = 2m$ and $b = 2n$ for some $m,n \in \mathbb{Z} $.   Thus  $$ a^2 = (2m)^2 \ \mbox{and}\ \  b^2= (2n)^2$$ $$ a ^2 + b^2= (2m)^2 + (2n)^2$$ $$ a^2 + b^2 = (4 m^2 + 4n^2) = 2(2m^2+2n^2) $$  Now $$2m ^ 2 + 2n^2\in \mathbb{Z} = k$$ Then  $$ a^2+b^2= 2k$$  Is this a correct proof?/ Any better ways to prove this.  Sorry for the lousy English. Thanks.
113249
How many positive integer $n$ are there  such that $2n+1$ , $3n+1$ are both perfect squares ?   $n=40$ is a solution . Is this the only solution ? Is it possible to tell whether finitely many or infinitely many solutions exist ?
57367
I know the following theorems:   A number can be represented as a sum of two squares precisely when $N$ is of the form $n^2 \prod p_i$ where each $p_i$ is a prime congruent to 1 mod 4 If the equation $a^2 + 1 \equiv a (\text {mod} p)$ is solvable for some $a$, then $p$ can be represented as a sum of two squares.  A positive integer $n$ is the sum of two squares iff each prime factor $p$ of $n$ such that $ p \equiv 3 (\text{mod} 4)$ occurs to an even power in the prime factorization of $n$   My problem is often which one to use and exactly how to use it.  Here is an example of a question:  Determine whether the following integers can be written as the sum of two squares. a) 363 b) 700 c) 34300 d) 325  For example, I figures 363 can not be written as a sum of two squares as it is of the form $4k +3$ where $k = 90$  For 700, which is not in the above stated form, I found $ 700 =2^2 \cdot 5^2 \cdot 7$ and 7 is the only factor $ p \equiv 3 mod 4$ and it is definitely not to an even power. Does this mean I'm done showing it is not the sum of two squares?  c) $34300 = 7^3 \cdot 5^2 \cdot 2^2$; again the power of 7 is not even.  d) Then I have $325 = 13 \cdot 5^2$ which has no prime factors congruent to 3 modulo 4. BUT $325 = 5^2(13)$ where $13 \equiv 1 mod 4$ and 5 is the $n$ mentioned in the first theorem. Is this enough to conclude that 325 is a sum of two squares?
90107
(1) For every $n\in \Bbb N$, there exist squarefree numbers $x$ and $y$ such that $x+y=n$, i.e., every natural number can be written as the sum of two squarefree numbers; (2) For every $n\in \Bbb N$, there exist squarefree numbers $k$ and $l$ such that $k-l=n$, i.e., every natural number can be written as the difference of two squarefree numbers.   For the squarefree numbers, I only know the definition of it, and I really did not quite understand how to prove these two from the definition.   Any help, please. Thanks in advance.
21724
A cube has numbers ±1 placed at each vertex and each center of cube’s six faces (so 14 numbers altogether). Moreover, it is known that each number in the center of a square face equals to the product of the 4 numbers at the vertices of this face. Can the sum of all 14 numbers be equal to zero? My first thoughts were test out some combinations and look for a pattern, but I was not able to notice anything major about it. So, the only progress I was able to really make was breaking it down into $V_1*V_2*V_3*V_4$, and so on for each face. Any help is greatly appreciated.
99288
Let $a \in \Bbb Z$ be such that $a = b^2 + c^2,$ where $b,c \in \Bbb Z \setminus \{0\}.$ Then $a$ cannot be written as $:$   $(1)$ $p d^2,$ where $d \in \Bbb Z$ and $p$ is a prime with $p \equiv 1\ \left (\text {mod}\ 4 \right ).$   $(2)$ $p q d^2,$ where $d \in \Bbb Z$ and $p,q$ are distinct primes with $p,q \equiv 3\ \left (\text {mod}\ 4 \right ).$  $(1)$ is false because $2^2 + 1^2 = 5 = 5 \cdot 1^2,$ where $d = 1 \in \Bbb Z$ and $5$ is a prime with $5 \equiv 1\ \left (\text {mod}\ 4 \right ).$ How do I prove or disprove the other option? Any help in this regard will be appreciated. Thanks for your time.
102378
There are four essentially different representations of $1885$ as the sum of squares of two positive integers. Find all of them.   I am guessing the must be some nice way of solving $z= x^2 + y^2 $  where x and y are integers in general that i don't know.    Is there some general way to know how many solution exists or bound it without computing it?
68102
I have thought a lot but am failing to arrive at anything encouraging.   First try: If this is to be proved by contradiction, then I start with the assumption that let $n$ be a number which is a sum of two numbers, of which at least one is prime. This gives $n = p + c$, where $p$ is the prime number and $c$ is the composite number. Also, any composite number can be written as a product of primes. So I can say, $n = p + p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$. From this, I get $n - p = p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$, but I have no clue what to do next.  Second try: For an instant let me forget about contradiction. Since $n &gt; 11$, I can say that $n \geq 12$. This means that either $p \geq 6$ or $c \geq 6$. Again I'm not sure what to do next.  Finally, consider that the number 20 can be expressed in three different ways: $17+3$ (both prime), $16+4$ (both composite), and $18+2$ (one prime and one composite). This makes me wonder what we are trying to prove.  The textbook contains a hint, "Can all three of $n-4$, $n-6$, $n-8$ be prime?", but I'm sure what's so special about $4, 6, 8$ here.
88083
Problem   Let $n = p_1.p_2.p_3 \cdots p_k.m^2$, where $p_1, p_2, p_3 \cdots p_k$ are distinct primes. Prove that n is sum of two squares if and only if $p_i$ is either 2 or $p_i \equiv 1 \pmod{4}$   For $p_i = 2$ , this is trivial case since $2m^2 = m^2 + m^2$. For $p_i \equiv 1 \pmod{4}$, I tried to use the fact that the product of a number of the form $4k + 1$ is also in this form. So I come up with: $$n = (4k + 1) \cdot m^2 = 4km^2 + m^2$$ Apparently, $m^2$ is a square, but I could not figure out how to prove $4km^2$ is a square, since k is in unknown form. Any idea?    Thanks,
89094
There exists a well-known result concerning a number of representations of $n$ as a sum of two squares. Is there anything similar for a number of representations of $n$ as a difference of two squares? Thank you.
22327
Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact?
100293
Related:    Except for the obvious cases $n=0,1$, are there any values of $n$ such that $\sum_{k=1}^n\sqrt k$ is an integer? How does one even approach such a problem? (This is not homework - just a problem I thought up.)
63842
Let $n\in\mathbb N$ and let $a_1,\ldots,a_n$ be natural numbers smaller than $10^n$. Write each $a_k$ in base $10$ and add $0$'s to the left of each decimal expansion, if needed, so that each $a_k$ is written with $n$ digits. Consider the matrix such that the entries of the $k$th row are the digits of $a_k$ (written in the same order). Let $d\in\mathbb N$ be such that $d$ divides each $a_k$. Prove that $d\mid\det A$.   For instance, suppose that $n=4$ and that your numbers are $3876$, $2784$, $684$, and $8388$, each of which is a multiple of $12$. Then$$A=\begin{bmatrix}3 &amp; 8 &amp; 7 &amp; 6 \\  2 &amp; 7 &amp; 8 &amp; 4 \\ 0 &amp; 6 &amp; 8 &amp; 4 \\  8 &amp; 3 &amp; 8 &amp; 8\end{bmatrix}$$and $\det A=-360$, which is, in fact, a multiple of $12$.  I learned about this problem yesterday. I found it quite cute and I decided to share it with all of you.  Note: I know how to prove it.