Prove that for ${a^3} = {b^2} + {c^2}$. where $a$, $b$ and $c$ are positive integers that there are an infinite number of values for $a$.
$\begingroup$
$\endgroup$
2
-
3$\begingroup$ Hint: If $a=u^2+v^2$ with $u,v$ non-zero integers then there are non-zero integers $b,c$ so that $a^3=b^2+c^2.$ $\endgroup$ – Thomas Andrews Nov 21 '19 at 19:00
-
$\begingroup$ I think I got an idea of how to prove it give me a sec. $\endgroup$ – Prince Deepthinker Nov 21 '19 at 19:00
Add a comment
|
$\begingroup$
$\endgroup$
11
Let suppose first that exist $u,v\in\mathbb{Z}$ such that $a=u^2+v^2$ then $a^{3}=(au)^2+(av)^2.$ That is, every number $a$ that is sum of two squares is solution to the problem. Now, $5=2^{2}+1^2$ so every number of the form $5n^2$ can be written as $(2n)^2+n^2$ so there we have infinite possible values to $a.$ For example, taking $n=1$ we have $5=2^2+1^2$ so $5^3=10^2+5^2,$ for $n=2$ we have $20=4^2+2^2$ so $20^3=80^2+40^2,$ etc...
-
$\begingroup$ If you don't mind can you like this question m so I can give you an upvote $\endgroup$ – GoD of SE II Nov 21 '19 at 19:36
-
-
-
$\begingroup$ OK, I think I write it badly: the thing is that, if you can write a number as sum of two squares then the cube of that number is also a sum of two squares. Then I pick a simple family of numbers that can be written as sum of two squares. I added that point to my answer. A more interesting problem would be add the condition that a,b and c be coprime $\endgroup$ – Nsn998 Nov 21 '19 at 19:40
-
$\begingroup$ You got a typo here 5 to the power of 5? $\endgroup$ – GoD of SE II Nov 21 '19 at 19:44
$\begingroup$
$\endgroup$
6
I have got one way of demonstrating it:
Where $z$ is an odd number
Where $n=(z-1)/2$
1). $(10^z)^3=(3•10^{z+n})^2+(10^{z+n})^2$
This means that $10^{3z}= 9•10^{3z-1}+10^{3z-1}$ for $3z-1=2z+2n$ since $n=(z-1)/2$
Therefore $1•10^{3z}=(9/10)10^{3z}+(1/10)10^{3z}$ which is also $10^{3z}(9/10+1/10)$ which is also $1•10^{3z}$
Therefore there are an infinite number of values for z
now if one superimposes what is in the brackets of 1). with $a$, $b$ and $c$ you will get $a^3=b^2+c^2$
Therefore $a$ has an infinity of values
-
$\begingroup$ Disclaimer I am not a mathematician $\endgroup$ – Prince Deepthinker Nov 21 '19 at 19:17
-
-
$\begingroup$ Have an upvote for attempting, I need work through this $\endgroup$ – GoD of SE II Nov 21 '19 at 19:22
-
$\begingroup$ It seems watertight but I want some concensus on behalf of the community before I give it the green tick $\endgroup$ – GoD of SE II Nov 21 '19 at 19:31
-
1$\begingroup$ I agree that this is correct (it's similar to the other answer, using $10=3^2+1^2$ rather than $5=2^2+1^2$ as its 'base case', sort of) but the exposition is a little bit rambling and makes for (IMHO) unclear reading. $\endgroup$ – Steven Stadnicki Nov 21 '19 at 19:53
$\begingroup$
$\endgroup$
Let $n$ be any odd positive integer. Then:
$$(2^n)^3 = 2^{3n} = 2 \times 2^{3n-1}= 2 \times (2^{(3n-1)/2})^2$$
The requirement that $n$ is odd ensures that $(3n-1)/2$ is an integer.
