If $x=a^2+b^2$ and $y=c^2+d^2$   how can i prove that xy is also the sum of two rational squares?   My teacher told me there are various methods to attack this problem but an easy way is to use the norms of guassian integers but I can't see how it would help.
122307
Given $a,b,c,d \in \mathbb{Z}$, there is $x,y$ such that  $$(a^2 + b^2)(c^2 + d^2) = x^2 + y^2$$  One can show this by considering the complex number $a + bi$ and $c+ di$, using complex properties to deduce that $x = ac - bd, y = ad + bc$ is a solution.   However, given that either $a^2, b^2$ are distinct or $c^2,d^2$ are distinct, then how can one can find nonzero solutions $x,y$?  Sorry, I am assuming $a,b,c,d$ are nonzero.
38131
So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$. By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+b(2axby+by^2)$. It doesn't help my proof. Please help me with this proof.
94710
Given an integer $m=pq$, where $p,q$ are both primes such that $p\equiv 1 \pmod{4}, q\equiv 1 \pmod{4}$. It is known that $p$ can be written as a sum of two squares (of positive integers) in a unique way, and the same for $q$. Prove that $m$ can be written as a sum of two squares (of positive integers) in exactly two distinct ways.  Attempt  Notice that for positive integers $u,v,A,B$, we have \begin{align*}(u^2+v^2)(A^2+B^2)&amp;=(uA+vB)^2+(vA-uB)^2=(vA+uB)^2+(uA-vB)^2\end{align*} Therefore if $p=a^2+b^2,q=c^2+d^2$, then \begin{align*}m&amp;=(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(bc-ad)^2=(bc+ad)^2+(ac-bd)^2\end{align*} meaning that $m$ can be written as a sum of two squares (of positive integers) in at least two distinct ways.  How do I prove that there doesn't exist a third way of writing $m$ as a sum of two squares (of positive integers)?
98111
Prove that $x^2+y^2=z^n$ has a solution $(x, y, z)$ in $\mathbb{N}$ for all $n\in\mathbb{N}$  I tried to prove this by induction, but couldn't. ( This was probably because the solution for some $n$ isn't necessarily related to the solution for $n+1$)  I can't seem to see any other way other than induction for proving the statement. Any help/hints on solving this problem, or any alternative approach will be highly appreciated :)
63677
let $a,b,c&gt;0$. Show that   $$a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$$   By the $$(1+a)(1+b)(1+c)=1+a+b+c+ab+bc+ac+abc$$ it is suffient to show  $$a^2+b^2+c^2+abc+2\ge a+b+c+ab+bc+ac$$ How?
74230
In Abbot's Understanding Analysis I am asked to show that $\sqrt{2}+\sqrt{3}$ is an algebraic number. I have shown that those two are algebraic separately (that was simple), but I can't figure out what to do to show that their sum is algebraic, too. For example, I tried $(\sqrt{2}+\sqrt{3})^{2}=5+\sqrt{24}$ and then I tried to think of a polynomial of form $ax^2-bx^{0}=0$ e.g. $c(\sqrt{2}+\sqrt{3})^{2}-c(5+\sqrt{24})=0$ that would work, but couldn't find the $c$ value that would make $b=c(5+\sqrt{24})$ integer, but couldn't find one. Maybe I can somehow conjecture that the sum of two algebraic numbers must be algebraic, too, but I was wondering if there's a way to find a polynomial to show this. Thanks!
31797
Prove that $$p(x,y) = x^4 + y^4 + x^2 + y^2$$ can be written as a sum of squares of three polynomials over $x,y$ for real numbers.
910
I stumbled across this relationship while I was messing around. What's the proof, and how do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to $2x + 1$ should equal $x^2$.
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
36411
Prove that, for all positive real numbers $x$, $y$ and $z$, $$4(x + y + z)^3 &gt; 27(x^2y + y^2z + z^2x)$$  I've tried expressing it as a sum of squares, but haven't got anywhere.   Hints are also welcome.
16824
How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?
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.
62095
If $x$ and $y$ are positive integers, then when is $2^x+3^y$ a perfect square?   I tried this question a lot but failed. I tried dividing cases into when $x,y$ are even/odd, but still have no idea what to do when they are both odd. I tried looking into when is  $2^x+3^y$ is of the form $6k+1$ and so on .. but couldn't succeed. Can anyone help?
44952
Let $a$, $b$, and $c$ be positive real numbers. Prove that  $$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$  Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?
102624
The question is : prove that $y^2 = x^3+(x+4)^2$ have no solutions in positive integers $x,y$.  I tried to play with the equation and get to $ x^3 = (y+x+4)(y-x-4)$, if $p|x$ then $p^3 | x^3$ and $ p | y+x+4$ or $ p | y-x-4$ , assume that $p$ divides both then we get that $ p | 2x-8$ and since $p|x$ we get that $ p | 8$ meaning $p=2$, so if we assume that $x$ is odd then $ p^3 | y+x+4$ or $p^3| y-x-4$ but not both and so we can write $y+x+4 = b^3 $ and $ y-x-4= a^3$, but that did not help me solving the problem.
55143
The problem's quite clear. Prove that $$\gcd(a^2, b^2) = \gcd(a, b)^2$$  This is easy to understand intuitively and using the Fundamental Theorem of Arithmetic would be easy but I want to prove it by using the Divisibility axioms and the GCD. I have attempted it and am mentioning my results.  Here $d = \gcd(a, b)$.   $$d | ax + by \implies d^2 | a^2x^2 + 2axby + b^2y^2 \implies d^2 | a^2x^2 + b^2y^2 \implies \gcd(a^2, b^2) | d^2$$  I feel that this is not necessary. Please help.  Also, the proof can be generalized to $$\gcd(a^n, b^n) = \gcd(a, b)^n$$  Any help would be appreciated.
73198
Show that any non-negative rational integer may be written as the sum of two squares if it is of the form $$2^ip_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}q_1^{2j_1}q_2^{2j_2}\cdots q_n^{2j_n}$$ where the $p_1,p_2,\dots,p_m$ are all primes congruent to $1\bmod 4$, the $q_1,q_2,\dots,q_n$ are all primes congruent to $3\bmod 4$ and $i,k_1,k_2,\dots,k_m,j_1,j_2,\dots,j_n$ are all non-negative integers.   I am completely blank on how to form a solution, so help would be much appreciated.