INFINITY SOLUTIONS
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Prove or disprove that, the equation $ x^2+y^2+z^2 =59^n $ has solution for every positive value of n.
- Phlembac Adib Hasan
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Re: INFINITY SOLUTIONS
Well here is my proof.
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Re: INFINITY SOLUTIONS
Let,n=2 ,so, the equation is $x^2+y^2+z^2=59^2$.sm.joty wrote:Well here is my proof.It's easy to see that $x,y,z$ must be odd.
Now, $59^2=(2.5^2+3^2)^2=2^2.5^4+3^4+2.2.5^2.3^2 $
Now,let,$ x^2=2^2.5^4,y^2=3^4,z^2=2.2.5^2.3^2 $ so,I don't understand why x,y,z must be odd?
- afif mansib ch
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Re: INFINITY SOLUTIONS
@mathpritom via,
\[59\equiv 3(mod4)\]
for n even and odd respectfully \[n^2\equiv o(mod4)\]
or\[n^2\equiv 1(mod4)\]
but\[x^2+y^2+z^2\equiv 3(mod4)\]
can only be possible when \[x^2,y^2,z^2\equiv 1(mod4)\]
so x,y,z are odd.
\[59\equiv 3(mod4)\]
for n even and odd respectfully \[n^2\equiv o(mod4)\]
or\[n^2\equiv 1(mod4)\]
but\[x^2+y^2+z^2\equiv 3(mod4)\]
can only be possible when \[x^2,y^2,z^2\equiv 1(mod4)\]
so x,y,z are odd.
- Phlembac Adib Hasan
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Re: INFINITY SOLUTIONS
@ Mansib vaia, this proof works for odd $n$s.But for even $n$, all of $x,y,z$ can't be odd as $59^{2k} \equiv 1(mod\; 4)$.Actually two of them must be even and the other will be odd.(Like $58,9,6$)That's why Pritom Vaia was confused.I think Joty Vaia should replace the sentence "It's easy to see that x;y;z must be odd" by "It's easy to see that for odd $n$s,x;y;z must be odd".
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Re: INFINITY SOLUTIONS
Humm...... got the mistake.Phlembac Adib Hasan wrote:@ Mansib vaia, this proof works for odd $n$s.But for even $n$, all of $x,y,z$ can't be odd as $59^{2k} \equiv 1(mod\; 4)$.Actually two of them must be even and the other will be odd.(Like $58,9,6$)That's why Pritom Vaia was confused.I think Joty Vaia should replace the sentence "It's easy to see that x;y;z must be odd" by "It's easy to see that for odd $n$s,x;y;z must be odd".
Actually I should replace that "two of the $x,y,z$ is even or all of them is odd."
But the think is that we can just use induction for both odd and even $n$
There is no need to use Modular.
So I should use induction.
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Re: INFINITY SOLUTIONS
Note that $59=5^2+5^2+3^2$. Then it immediately follows from the known fact that The product of two numbers of the form $a^2+b^2+c^2$ is of the same form.(Prove yourself)MATHPRITOM wrote:Prove or disprove that, the equation $ x^2+y^2+z^2 =59^n $ has solution for every positive value of n.
By the lemma, we can conclude that any power of $5^2+5^2+3^2$ is of the same form $x^2+y^2+z^2$. Thus proved.
One one thing is neutral in the universe, that is $0$.