 # Show that the congruence 8x^2 -x + 4 -= 0(mod 9) has no solution by reducing to the form y^2 -= a(mod 9) nicekikah 2021-03-09 Answered
Show that the congruence $8{x}^{2}-x+4\equiv 0\left(\text{mod}9\right)$ has no solution by reducing to the form ${y}^{2}\equiv a\left(\text{mod}9\right)$
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Step 1
Given congruence is
$8{x}^{2}-x+4\equiv 0\left(\text{mod}9\right)$
We have to show this congruence has no solution by reducing it to the form ${y}^{2}=a\left(\text{mod}9\right)$
Step 2
Consider the congruence
$8{x}^{2}-x+4\equiv 0\left(\text{mod}9\right)$
$⇒-{x}^{2}+8x+4\equiv 0\left(\text{mod}9\right)\because -1\equiv 8\left(\text{mod}9\right)$
$⇒-{x}^{2}+8x-16+16+4\equiv 0\left(\text{mod}9\right)$ by adding and substracting 15
$⇒-{x}^{2}+8x-16+20\equiv 0\left(\text{mod}9\right)$
$⇒-\left({x}^{2}-8x+15\right)+20\equiv 0\left(\text{mod}9\right)$
$⇒-{\left(x-4\right)}^{2}+20\equiv 0\left(\text{mod}9\right)$
$⇒-{\left(x-4\right)}^{2}\equiv 20\left(\text{mod}9\right)$
$⇒{\left(x-4\right)}^{2}\equiv 20\left(\text{mod}9\right)\because gcd\left(-1,9\right)=1$, so cancellation law holds
$⇒{\left(x-4\right)}^{2}\equiv 2\left(\text{mod}9\right):0\equiv 2\left(\text{mod}9\right)$
Take y = x-4, the we have the congruence
${y}^{2}\equiv 2\left(\text{mod}9\right)$
and the congruence ${y}^{2}\equiv 2\left(\text{mod}9\right)$ has no solution.
$⇒$The given congruence has no solution.