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Dealing with Diophantine equation I saw the following to be true and could arrive at a proof. Has this been dealt with earlier any where ? $$x^p+y^q=z^r $$, where $$x, y, p, q, z, r$$ are all natural numbers. It can be shown that the above statement is never true when $p $ and $q$ are even integers and $x$ and $y$ are odd integers and $r > 1$

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  • $\begingroup$ Yes, a modulo four consideration gives this immediately. Except when $x=y=r=1$ and $z=2$. $\endgroup$ – Jyrki Lahtonen May 2 '17 at 5:24
  • $\begingroup$ @JyrkiLahtonen Can you please elaborate ? $\endgroup$ – Barun Dasgupta May 2 '17 at 10:14
  • $\begingroup$ If $x$ and $y$ are both odd, and $p$ and $q$ are both even, then $x^p+y^q\equiv2\pmod 4$. This forces $z$ to be even, but unless $r=1$ $z^r$ is then divisible by $4$. So $r=1$. But if we allow $r=1$, then the existence of solutions is not interesting. Sorry about exaggerating a bit in the first comment. It is just that I easily deduced $r=1$ and lost interest. $\endgroup$ – Jyrki Lahtonen May 2 '17 at 10:38
  • $\begingroup$ @JyrkiLahtonen Yes by mistake I forgot to say that r > 1, thanks . It is interesting to note that when p= q=r a very small subset of FLT is proven with simple mathematics, which has been proven by the method of infinite descent. $\endgroup$ – Barun Dasgupta May 2 '17 at 11:18
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Without that requirement, there are solutions such as $$ \eqalign{0^2 + 1^3 &= 1^2\cr 1^2 + 2^3 &= 3^2\cr 3^2 + 3^3 &= 6^2\cr 6^2 + 4^3 &= 10^2\cr 10^2 + 5^3 &= 15^2\cr 15^2 + 6^3 &= 21^2\cr 21^2 + 7^3 &= 28^2\cr 28^2 + 8^3 &= 36^2\cr 36^2 + 9^3 &= 45^2\cr 45^2 + 10^3 &= 55^2\cr \ldots&\cr }$$

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  • $\begingroup$ How do you generate these solutions from scratch? $\endgroup$ – Toby Mak May 2 '17 at 6:58
  • $\begingroup$ @Robert Israel . You mean to say when the restriction of q being even is removed ? $\endgroup$ – Barun Dasgupta May 2 '17 at 7:26
  • $\begingroup$ @TobyMak $\left(\frac{y^2-y}2\right)^2 + y^3 = \left(\frac{y^2+y}2\right)^2$. $\endgroup$ – Robert Israel May 2 '17 at 15:23

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