Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find all positive integer $x,y$ such $$5xy\sqrt{(x^2+y^2)^3}$$ can write the sum of Four 5-th powers of positive integers.In other words: there exst $a,b,c,d\in N^{+}$ such $$5xy\sqrt{(x^2+y^2)^3}=a^5+b^5+c^5+d^5$$

This problem is from Math competition simulation test.I seach this problem and found this problem background is Euler's sum of powers conjecture.can see link

maybe this problem is not hard.because is from competition. since $$5xy\sqrt{(x^2+y^2)^3}=5xy(x^2+y^2)\sqrt{x^2+y^2}$$ so $$x^2+y^2=m^2$$ $$x=3,y=4,m=5$$ and $$x=(a'^2-b'^2),y=2a'b',m=a'^2+b'^2$$ then I can't it Thank you for you help .

share|improve this question
1  
Well,first of all,$\sqrt{x^2+y^2}$ must be an integer. –  rah4927 Apr 19 at 4:44
    
They look suspiciously like pythagorean triplets,don't they?Oh wait. . . . –  rah4927 Apr 19 at 4:51
    
Note that $a$ and $b$ must have opposite parity.It isn't immediately useful,but keeping this information in mind will prevent making any errors in the future.Besides,have you actually tried and substitue the value of the variables $x,y$(in terms of $a,b$) in the original equation? –  rah4927 Apr 19 at 5:28
    
Substitute the values of x and y in the original equation and then try and see what you can get.It will probably help.Also,since our equation is symmetric in x and y,assume without loss of generality that $a$ is odd and then try parity.I am not saying this will yield the solutions,but it might help. –  rah4927 Apr 19 at 8:53
1  
"maybe this problem is not hard.It's from a competition".haha,contest problems aren't hard? –  rah4927 Apr 19 at 9:19

1 Answer 1

Something that might help. Using what you said, namely:

$x^2+y^2=k^2$, with $x=m^2-n^2$, $y=2mn$ then the left hand side of the equation becomes: $LHS=10(m^9n-mn^9+2m^7n^3-2m^3n^7)$

Since $RHS\equiv 0$ modulo 2 and modulo 5 then we have $a+b+c+d\equiv 0 \mod{2}$ and $a+b+c+d\equiv 0 \mod{5}$

share|improve this answer
    
Note that "odd modulo 5" makes no sense, i.e. $(0, 0, 1, 4)\equiv (0, 0, 1, 9)\pmod{5}$, the latter being acceptable $\pmod{10}$. –  jpvee Apr 29 at 14:02
    
@jpvee Thank you. I removed that part. –  Test123 Apr 29 at 15:03
    
@Test,I don't understand how you derive the final line of your ans. –  rah4927 Apr 29 at 15:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.