establishes that there is no integer $a$ to which it satisfies $a^{2}+3a+5\equiv0 \pmod {121}$
Asked
Active
Viewed 54 times
-1
-
1Satisfies what? For any $a \in \mathbb{Z}$ is $a^2 + 3a + 5 \in \mathbb{Z},$ therefore can be a remainder. You meant $a^2 + 3a + 5 \equiv 0 \pmod {121}?$ – 1b3b May 19 '20 at 22:35
-
1sorry for this. – Brayan Alexis Benitez May 19 '20 at 22:43
-
Maybe this would be helpful..$a^2 + 3a + 5 \equiv 0 \pmod {121} \iff$ $a(a + 3) \equiv - 5 \equiv 116 \pmod {121}$ and now it's easy to see that there is no solution. – 1b3b May 19 '20 at 22:46
-
If $a^2+3a+5\equiv a^2-8a+16=(a-4)^2\equiv0\bmod11$ then $11|a-4$ – J. W. Tanner May 19 '20 at 22:50
-
@J.W.Tanner OP will need a bit of "lifting" to $11^2$ as well. – Anurag A May 19 '20 at 22:53
2 Answers
0
The discriminant is $-11$ which should be a big hint how to get an easy proof;
$a^2 + 3a + 5 \equiv 0 \pmod {121}$ is equivalent to $4a^2 + 12a + 20 \equiv 0 \pmod {121}$ or $(2a+3)^2+11 \equiv 0 \pmod {121}$
But now we get $2a+3$ divisible by $11$ hence $(2a+3)^2$ divisible by $121$ and that contradicts the above since $11$ is not divisble by $121$. Done!
Conrad
- 27,433
0
Idea: Whenever I see a congruence problem involving quadratics, I always complete squares, to make the equation simpler. If you have some expression like $S=mx^2+nx$, I multiply it by $4m$, first, so that $4mS=4m^2x^2+4mnx=(2mx+n)^2-n^2$.
Solution Suppose we have $a^2+3a+5\equiv 0\mod 121$. We multiply the equation by $4$ to complete squares: $4a^2+12a+20\equiv 0\mod 121$. Then, we get the equation: $$(2a+3)^2+11\equiv 0\mod 121.\,\,\,\,\,\,\,(1)$$ Note, that looking mod $11$, we get that $(2a+3)^2\equiv 0\mod 11$. Which means that $(2a+3)^2$ is a square that is a multiple of 11. iIn particular, $(2a+3)^2$ has to be a multiple of $11^2=121$, i.e. $(2a+3)^2\equiv 0\mod 121$. This is a contradiction to the equation (1) above.
Julian Mejia
- 4,438