Let $x, y$, and $z$ be integers with $z>1$. Show that \[(x+1)^{2}+(x+2)^{2}+\cdots+(x+99)^{2}\neq y^{z}.\]
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Tags: Diophantine Equations, pen
TTsphn
21.10.2007 07:18
Peter wrote: Let $ x, y$, and $ z$ be integers with $ z > 1$. Show that \[ (x + 1)^{2} + (x + 2)^{2} + \cdots + (x + 99)^{2}\neq y^{z}. \] $ 99x^2+2x(1+...+99)+(1^2+..+99^2)=y^z$ $ \Rightarrow 3|y$ imply that $ 9|y^z$ But $ 9\not |99x^2+2x(1+...+99)+(1^2+..+99^2)$ So it hasn't solution.
reveryu
08.05.2017 17:18
let $x=t-50$ ,we have $99t^2+2*25*33*49=y^z$ clearly $v_2(y^z)=0$ or $1$ so $z=1$, contradiction
Wave-Particle
26.07.2017 05:15
Let $k = x + 45$. We wish to show $$(k - 44)^2 + (k-43)^2 + \cdots + (k+44)^2 = y^z$$has no solutions. Once the terms on the LHS cancel, we are left with $$99k^2 + 2\cdot \sum_{i=1}^{44} i^2 = 99k^2 + 44\cdot 15\cdot 89 = y^z.$$Clearly by taking $\pmod 3$ on the equation, we get that $y\equiv 0\pmod 3$. However, taking the equation under $\pmod{9}$, we get that $y^z\equiv 0\pmod 9$ since $z>1$ but $99k^2+44\cdot 15\cdot 89\not\equiv 0\pmod 9$, a contradiction.
nangtoando
07.10.2017 06:02
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