# 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 $a$. [closed]

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 $$a$$.

• Hint: If $a=u^2+v^2$ with $u,v$ non-zero integers then there are non-zero integers $b,c$ so that $a^3=b^2+c^2.$ – Thomas Andrews Nov 21 '19 at 19:00
• I think I got an idea of how to prove it give me a sec. – Prince Deepthinker Nov 21 '19 at 19:00

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...

• If you don't mind can you like this question m so I can give you an upvote – GoD of SE II Nov 21 '19 at 19:36
• I dont follow you – GoD of SE II Nov 21 '19 at 19:37
• What does the first part mean? – GoD of SE II Nov 21 '19 at 19:38
• 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 – Nsn998 Nov 21 '19 at 19:40
• You got a typo here 5 to the power of 5? – GoD of SE II Nov 21 '19 at 19:44

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

• Disclaimer I am not a mathematician – Prince Deepthinker Nov 21 '19 at 19:17
• Hmm interesting – GoD of SE II Nov 21 '19 at 19:21
• Have an upvote for attempting, I need work through this – GoD of SE II Nov 21 '19 at 19:22
• It seems watertight but I want some concensus on behalf of the community before I give it the green tick – GoD of SE II Nov 21 '19 at 19:31
• 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. – Steven Stadnicki Nov 21 '19 at 19:53

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.