منتديات الرياضيات العربية - عرض مشاركة واحدة

02-03-2009, 10:59 AM

هذي محاولة بسيطة :

Find all integers $x,y$ satisfying

$x^2 + 300x + 9950 = 33y^2$

Solution

Take both sides given in the problem statement modulo 33

$x^2 + 3x + 17 \equiv{0} \pmod{33}$

but this last congruence is insolvable for all integers $x$ , to see this we could plug in values from $\lbrace 0,1,2,...,32 \rbrace$ for $x$ and check that none of them satisfy what we need. We could argue similarly ,and in a faster way, by taking it modulo 11

$x^2 + 3x + 6 \equiv{0} \pmod{11}$

and this congruence is insolvable either, b/c

$0^2 + 3 \cdot 0 + 6 \equiv 6 \neq 0 \pmod{11}$
$1^2 + 3 \cdot 1 + 6 \equiv 10 \neq 0 \pmod{11}$
$2^2 + 3 \cdot 2 + 6 \equiv 5\neq 0 \pmod{11}$
$3^2 + 3 \cdot 3 + 6 \equiv 2 \neq 0\pmod{11}$
$4^2 + 3 \cdot 4 + 6 \equiv 1\neq 0 \pmod{11}$
$5^2 + 3 \cdot 5 + 6 \equiv 2 \neq 0\pmod{11}$
$6^2 + 3 \cdot 6 + 6 \equiv 5\neq 0 \pmod{11}$
$7^2 + 3 \cdot 7 + 6 \equiv 10\neq 0 \pmod{11}$
$8^2 + 3 \cdot 8 + 6 \equiv 6\neq 0 \pmod{11}$
$9^2 + 3 \cdot 9 + 6 \equiv 4 \neq 0 \pmod{11}$
$10^2 + 3 \cdot 10 + 6 \equiv 4 \neq 0 \pmod{11}$

so the equation

$x^2 + 300x + 9950 = 33y^2$

has no integral solutions $x,y$ . END

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