Prove that for any integers $a, b, c, d$ number $(a^2 + b^2)(c^2 + d^2)$ is a sum of two squares of integer.  In fact I have no idea how to do this and I'll appreciate any tips or the  solution.
122526
Prove that if n is the product of two numbers that can be written as the sum of two squares then n can be written as the sum of two squares.
52355
For $a,b,c,d \in \mathbb{R}$, is $\sqrt{a^2+b^2} + \sqrt{c^2+d^2} \geq \sqrt{(a+c)^2 + (b+d)^2}$? I have tried to show this using the fact that $\sqrt{x+y}\leq\sqrt{x}+\sqrt{y}\; $  for non-negative $x,y$ but haven't been able to arrive at a proof.
26605
Prove that the sum of the digits of a perfect square can't be 2, 3, 5   , 6, or 8.   I'm completely stumped on this one, how would I go about proving it?
102108
Could anybody help me some ideas on the below problem:  Let $n$ be a positive integer. Prove that there are infinite pairs of positive integer $(x\, y)$ such that $$x^2+x+1=(y^2+y+1)(n^2+n+1).$$ Thanks in advance.
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,
22537
I was reasonably certain I've seen this before, but I was wondering how to solve the Diophantine equation  $$a^2+b^2=c^2+d^2$$  I tried a web search and found nothing on this one.  I'm trying to avoid another library trip to a less than local library (maybe I should have taken better notes on that chapter...).  I'm not quite sure how to handle this one.  The only thing I can figure out with this equation, if I remember correctly, is that the sum on either side may only contain prime factors of 2 or odd primes congruent to 1 mod 4.  And if I don't want a and b equal to c and d, the sum can't be prime as I believe a prime congruent to 1 mod 4 can be represented as the sum of 2 squares in exactly one way.  But that doesn't give me any insight into actually solving this problem.
2457
Let $a, b, c\geq 0$ s.t.$$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1$$  Show that $ a^2b^2+a^2c^2+b^2c^2\leq 3$.  My idea:  I denote $a+b+c=x$, $ab+bc+ac=y$ and $abc=z$.  Then I have $$x^2+y^2+z^2-xy-xz-yz+2z-y-x=0$$ I have to show that $$y^2-2xz\leq 3$$ I tried to prove it with the sign of trinom, but it doesn't work.
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.
44996
i would like to know how to prove that if $a^2+b^2=c^2$ then   a) $3|a$ or $3|b$  b) $5|a$ or $5|b$ or $5|c$  c) $4|a$ or $4|b$   Also that $a^2+b^2+c^2+1$ can't be divisible by 8  Thank you very much!
87485
Is it true that all numbers of the form $\sqrt[b_1]{a_1}+\sqrt[b_2]{a_2}+\cdots+\sqrt[b_k]{a_k}$ are not integers, where $a_i,b_i$ are positive integers such that none of the $k$ terms is an integer? I think this must be a theorem somewhere, probably a fundamental theorem in number theory, but not sure what it would be called.
10320
If $a$ and $b$ are two relatively prime positive integers such that $ab$ is a square, then $a$ and $b$ are squares.   I need to prove this statement, so I would like someone to critique my proof. Thanks  Since $ab$ is a square, the exponent of every prime in the prime factorization of $ab$ must be even.   Since $a$ and $b$ are coprime, they share no prime factors.   Therefore, the exponent of every prime in the factorization of $a$ (and $b$) are even, which means $a$ and $b$ are squares.
27965
Let $a$ and $b$ be two integers. Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$.   This is what I have done so far:  Let $d = \gcd(a,b)$. Then $d=ax+by$ for some $x,y$. Then $d^2 =(ax+by)^2 = a^2x^2 + 2axby+b^2y^2$.  I am trying to create a linear combination of $a^2$ and $b^2$ but do not know what to do with the middle term.  EDIT: I would be interested in seeing a proof using the Fundamental Theorem of Arithmetic as well. I simply thought it would be easiest to use Bezout's Identity.
23149
For any natural number $n\ge 1$, given pairs $(a_1,b_1),(a_2,b_2),...,(a_n,b_n)$ of integer numbers, there exist integer number $c$ and $d$ such that  $$\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$$    My initial approach is  Base Case: $(a_1^2+b_1^2) = a_1^2+b_1^2$ which is true. (Although it is trivial)  Prove the statement is true when $n=2$: We have $$(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$$ (Thanks André Nicolas for pointing it out)   So if $a,b,c,d$ are integers, $ac,bd,ad,bc$ are all integers and integers are closed under addition and subtraction. Hence $(ac-bd),(ad+bc)$ are integers.   Inductive Hypothesis: $\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$ is true  Inductive Step: $$\prod_{i=1}^{n+1}(a_i^2+b_i^2) = \prod_{i=1}^{n}(a_i^2+b_i^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2)$$  Where $c$ and $d$ are integers.   But when we apply $n=2$, we have  $(c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (e^2 + f^2)$ where $e$ and $f$ are integers.  Hence, by the principle of induction, the statement we needed to prove is true.
30934
It is easy to find 3 squares (of integers) in arithmetic progression. For example, $1^2,5^2,7^2$.  I've been told Fermat proved that there are no progressions of length 4 in the squares. Do you know of a proof of this result?   (Additionally, are there similar results for cubes, 4th powers, etc? If so, what would be a good reference for this type of material?)    Edit, March 30, 2012: The following  in MO is related and may be useful to people interested in the question I posted here.
25854
Prove that a perfect square is either a multiple of $4$ or of the form $4q+1$ for some $q\in \mathbb{Z}$.  Any ideas on how to start? Do I use a proof by contraposition? Also what's the definition of a perfect square?
103821
I saw this problem recently and found an elegant solution to it, and was curious to see if anybody would think of something else. Nice solutions to nice problems are fun to see!  Problem: Prove that, for all non-negative integers $k$, there exist integers $a$, $b$ such that $a^2+b^2=5^k$. Bonus: Prove that there exist such $a$ and $b$ that furthermore satisfy $(a,b)=1$.
59853
My question is about proving that if $$a^2 + b^2 = c^2,$$ then 5 divides abc. a,b,c are positive integers.