Difference between revisions of "1965 AHSME Problems/Problem 40"

Revision as of 23:41, 21 October 2021

Problem

Let $n$ be the number of integer values of $x$ such that $P = x^4 + 6x^3 + 11x^2 + 3x + 31$ is the square of an integer. Then $n$ is:

$\textbf{(A)}\ 4 \qquad \textbf{(B) }\ 3 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 1 \qquad \textbf{(E) }\ 0$

Solution

First, we wish to factor $P$ into a more manageable form. From the beginning of $P$ , we notice $x^4+6x^3$ , which gives us the idea to use $(x^2+3x)^2=x^4+6x^3+9x^2$ .

This gives us $P=(x^2+3x)^2+2x^2+3x+31$ This is not useful, but it gives us a place to start from.

We can then try $(x^2+3x+1)=x^4+6x^3+11x^2+6x+1$ . $P=(x^2+3x)^2-3(x-10)$ This is much more useful, as it moves all non-linear terms inside of a squared expression.

We can then say $P=(x^2+3x+1)^2-3(x-10)=a^2$ , where $a^2$ is the square of an integer mentioned on the problem. Right from here, we can set $x=10$ , which cancels out the $3(x-10)$ , giving $(100+30+1)^2=a^2$ . This gives us one solution, $x=10$ , $a=131$ .

We can then rearrange the expression, giving us $(x^2+3x+1)^2-a^2=3(x-10)$ . Factoring using difference of squares, we obtain $(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)$

We can then state that when $x$ is greater than $3$ and less than $-9$ , $x^2+3x+1$ will be greater than $|3x-10|$ . This is obtained by setting $x^2+3x+1>|3x-10|$ and then solving the inequality. We can then conclude that $x^2+3x+1+|a|>|3x-10|$ .

Next, we claim that $x^2+3x+1-|a| \geq 1$ or $x^2+3x+1-|a| \leq -1$ when $x \neq 10$ . We can prove this by first noting that since $x$ and $a$ are integers, $x^2+3x+1-|a|$ is an integer. Next, we shall assume that $x^2+3x+1-|a|=0$ . Solving this and plugging back into the original equation, we obtain $(2|a|)(0)=3(x-10)$ . Solving we obtain $x=10$ , which is a contraction to $x \neq 10$ . Therefore, $x^2+3x+1-|a| \neq 0$ and $x^2+3x+1-|a| \geq 1$ or $x^2+3x+1-|a| \leq -1$ .

Finally, we can go back to the equation $(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)$ We note that since $(x^2+3x+1+|a|)$ is larger than $3(x-10)$ , in order for there to be solutions, $(x^2+3x+1-|a|)$ must be in the range $(-1,1)$ . However, this contradicts what was proven earlier, so when $x \neq 10$ and $x < -9$ or $x > 3$ , there are no solutions for $x$ .

Now, all that remains to be checked are values of $x$ between $-9$ and $3$ . Using brute force and checking each value individually, we can assert that there are no such solutions for $x$ , leaving us with only $1$ solution, $x=10$ . Therefore, the answer is $\boxed{(D) }$

Solution by treetor10145

@@ Line 20: / Line 20: @@
 We can then try <math>(x^2+3x+1)=x^4+6x^3+11x^2+6x+1</math>.
-<cmath>P=(x^2+3x)^2-3(x+10)</cmath>
+<cmath>P=(x^2+3x)^2-3(x-10)</cmath>
 This is much more useful, as it moves all non-linear terms inside of a squared expression.

1965 AHSC (Problems • Answer Key • Resources)
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Difference between revisions of "1965 AHSME Problems/Problem 40"

Revision as of 23:41, 21 October 2021

Problem

Solution

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