Four different positive integers $a, b, c, d$ are such that $a^2 + b^2 = c^2 + d^2$. What is the smallest possible value of $abcd$?  $$a^2 - c^2 = d^2 - b^2$$  $$(a-c)(a+c) = (d-b)(d+b)$$  $$(a-c)(a+c) - (d-b)(d+b) = 0$$  So there is one pair I see:  $$a -c = d - b, a + c = d + b$$  Suppose WLOG, $a &gt; b &gt; c &gt; d &gt; 0$  But that doesnt strike anything.
124171
Four different positive integers $a, b, c$, and $d$ are such that $a^2 + b^2 = c^2 + d^2$ What is the smallest possible value of $abcd$?  I just need a few hints, nothing else. How should I begin?  Number theory?
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?
92378
As suggested on Mathoverflow () I am transfering this question to math-stackexchange:  I am looking for integer solutions of the following algebraic equations:  $ 2a^{2} + 2b^{2} - c^{2} - d^{2} = 0 $ Preferentially the solutions should obey $a+b+c+d=0$.   By inspection I found the following solutions: $(a,b,c,d)=(1,0,1,1)$ , $(a,b,c,d)=(0,1,1,1)$, $(a,b,c,d)=(1,1,-2,0)$  and $(a,b,c,d)=(1,1,0,-2)$.   Additional solutions can be generated by swapping the sign of $a,b,c,d$ or  by scaling the solutions that I have given by an integer.   As a theoretical physicist I am rarely working with these diophantine equations and hence I am wondering what other solutions of the equation exist that I have not taken into account.   I am looking forward to your responses.
25329
Find the least value of $\dfrac {3a}{b+c} + \dfrac{4b}{a+c}+ \dfrac{5c}{a+b}$ for positive $a, b, c$.    I tried using the Cauchy-Schwarz inequality, but could not proceed after a bad equation which couldn't be further solved .
71679
If $a,b,c$ are positive , show that $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$$  Trial: Here  I proceed in this way $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge  \dfrac{2bc}{b+c}+\dfrac{2ca}{c+a}+\dfrac{2ab}{a+b}$$ then how I proceed. Please help.
92549
Problem: For what pair of positive integers $(a,b)$ is $3^a + 7^b$ a perfect square.  First obviously $(1,0)$ works since $4$ is a perfect square, $(0,0)$ does not work, and $(0,1)$ does not work, so we can exclude cases where $a$ or $b$ are zero for the remainder of this post.  I have a few observations made but not much room for a full solution.   First, since powers of an odd number are odd, and the sum of two odd numbers is even, so the base of the square must be an even number.  Secondly, the last digit of the powers of $3$ are $\{1,3,7,9 \}$ , whereas the last digits of the powers of $7$ are $\{7,9,3,1 \}$.  From here I am not sure how to proceed and any hints much appreciated. I'm not sure if there a finite amount of pairs or not either.
29391
find all  positive integers $a,b,c$ such that  $abc=24$  $ab+bc+ca=38$  If particular values are given then we can easily find the solution but I am searching for some short general method.  Is there any sufficient condition for the constant values to have integer solution?
108457
Let $a_1=10$, $a_2=20$ and define $$a_{n+1}=a_{n-1}-\frac{4}{a_n}$$ for $n \gt 1$. What is the smallest value of $k$ for which $a_k=0?$
39384
Let $a, b$ and $c$ be positive real numbers. Prove that   $$a^2 b + b^2 c + c^2 a + ab^2 + bc^2 + ca^2 \ge 6abc$$   I have tried factoring with no success for the better part of an hour or so.  How can I go about this?
68034
If $a$,$b$ and $c$ positive real numbers such that $a+b+c=1$, prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$. I have tried several methods to solve this,but can't get any result. Any idea?
29905
I'm reading a proof in my number theory textbook that all primes of the form $p = 4k+1$ are uniquely the sum of two squares. I'm stuck right at the beginning of the proof, where they say:  To establish the assertion, suppose that $$ p = a^2 + b^2 = c^2 + d^2 $$ where $a,b,c,d$ are all positive integers. Then $$ a^2 d^2 - b^2 c^2 = p(d^2 - b^2). $$  Perhaps I'm just missing something obvious, but I can't figure out how they managed to conclude that $a^2 d^2 - b^2 c^2 = p(d^2 - b^2).$ Please advise.
90003
I have to find an infinite set for $a,b,c,d$ such that $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb N$  I have found one possible set which is $x,2x,2x,3x$ and $x\in\mathbb N$  But I want another infinite sets and how to reach them because I have found my solution by hit and trial .
41594
I want to prove $gcd(a, a+2)$ = 1 or 2 for every integer $a$.   I know that (a, a+2) is equal to the least positive value of $ax + (a+2)y$, where $x, y \in \mathbb{Z}$. However, I'm a little stuck at this point. I'm thinking I should split apart the problem into two cases, one where $a$ is odd and one where $a$ is even? Any suggestions? Thanks.
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!
34036
Prove for positive integers $a,b,c$ and $d$  (where $b  \neq 0$ , $d \neq 0$ and $b \neq d$), if $\gcd(a,b) = \gcd(c,d) = 1$, then $\frac{a}{b} + \frac {c}{d}$ is not an integer.  I understand that if $\gcd(a,b)$ and $\gcd(c,d) = 1$, at least one number in each pair is a prime or is $1$. As for after that, I'm totally stumped, could I get some tips, clues, help?
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.
95047
Let $a, b$ and $c$ three positive real numbers such that $a+b+c=3$. Find the minimum value of $$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}.$$  Here is my attempt. By symmetry we can assume that $a\leq b\leq c$. The function $f(x)=\frac{2-x^3}{x}$ is decreasing and convex on $]0,2^{1/3}]$. So if $c\leq 2^{1/3}$ then $3f(1)=3f(\frac{a+b+c}{3})\leq f(a)+f(b)+f(c)=A$ with equality if $a=b=c=1$.   If $c&gt;2^{1/3}$ I don't see how to proceed.