Question

Show that there are infinitely many systems of positive integers (x,y,z,t) which have no common divisor greater that 1 and such that
$x^3+y^3+z^2=t^4.$

• A. $t=b^3+1$ have no common divisor greater than 1.
• B. $t=b^3-6$ have no common divisor greater than 1.
• C. $t=b^2+1$ have no common divisor greater than 1.
• D. $t=b^1+3$ have no common divisor greater than 1.

Answer 1

The correct option is A $t=b^3+1$ have no common divisor greater than 1.

Consider the identity $(a+1{)}^4-(a-1{)}^4=8a^3+8^a$. Taking $a=b^3$, with b an even integer gives $(b^3+1{)}^4={\left(2b^3\right)}^3+{\left(2b\right)}^3+((b^{3}1{)}^2{)}^2$ Since b is even,$b^3+1$ and $b^{3}1$ are odd integers. It follows that the numbers $x=2b^3,y=2b,z=(b^{3}1{)}^2$ and $t=b^3+1$ have no common divisor greater than 1.