integer solutions

by hemangsarkar, Jul 26, 2013, 6:48 AM

Q) Determine, with proof, the number of integer triples (x, y, z) satisfying
x^2 + 2y^2 + 98z^2 = 77777....7777 (2013 digits)

my solution :

working modulo $7$,

$x^2 + 2y^2 \equiv 0\pmod7$

all perfect squares are $(0,1,2,4) \pmod7$

using all these cases - the only possibility is that

$x^2 \equiv 0\pmod7$ and $y^2 \equiv 0\pmod7$

that means : $x = 7m$ and $y = 7n$ for some integers $m,n$.

cancelling $7$ from both the sides - we have

$7(m^2 + n^2 + 2z^2) = \frac{10^{2013} - 1}{9}$

that means that $\frac{10^{2013} - 1}{9} \equiv 0\pmod7$

but we can see that by euler's totient theorem :

$10^6 \equiv 1\pmod7$

$10^{2010} \equiv 1\pmod7$

$10^{2013} \equiv 1000 \equiv 6\pmod7$

so, $10^{2013} - 1 \equiv 5\pmod7$

no integer solutions exist.
This post has been edited 1 time. Last edited by hemangsarkar, Jul 26, 2013, 6:50 AM

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