this seems to be very hard...

amparvardi wrote: Prove that the equation $x^3+y^3+z^3=t^4$ has infinitely many solutions in positive integers such that $\gcd(x,y,z,t)=1$. Amir,I think The original question is $x^3+y^3+z^2=t^4.$And In this problems two of my favourite parametres are $m+1,m-1$.And our intution says to take $z=(m-1)^2,t=(k+1)$.Then We get $x^3+y^3=8k^3+8k,$so let's take $k=p^3,$Now using $Gcd(x,y,z,t)=1,$ we easily find $p$ even.And this is enough to prove the claim

Amir Hossein wrote: Prove that the equation $x^3+y^3+z^3=t^4$ has infinitely many solutions in positive integers such that $\gcd(x,y,z,t)=1$. Mihai Pitticari & Sorin Rǎdulescu As we've $(n-1)^3+n^3+(\frac {(n-1)(n-2)}{2})^2=(\frac {(n)(n+1)}{2})^2$ So now it's enough to make $RHS$ a square. Which is trivial from Pell's equation. And also our numbers satisfying the gcd condition. So done.

Problem condition $x^3+y^3+z^3=t^4$ or $x^3+y^3+z^2=t^4.$?

lazizbek42 wrote: Problem condition $x^3+y^3+z^3=t^4$ or $x^3+y^3+z^2=t^4.$? https://artofproblemsolving.com/community/c3046h1048092