Difference between revisions of "1965 AHSME Problems/Problem 40"
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We can then try <math>(x^2+3x+1)=x^4+6x^3+11x^2+6x+1</math>. | 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+1)^2-3(x-10)</cmath> |
This is much more useful, as it moves all non-linear terms inside of a squared expression. | This is much more useful, as it moves all non-linear terms inside of a squared expression. | ||
Latest revision as of 00:42, 22 October 2021
Problem
Let be the number of integer values of
such that
is the square of an integer. Then
is:
Solution
First, we wish to factor into a more manageable form.
From the beginning of
, we notice
, which gives us the idea to use
.
This gives us
This is not useful, but it gives us a place to start from.
We can then try .
This is much more useful, as it moves all non-linear terms inside of a squared expression.
We can then say , where
is the square of an integer mentioned on the problem. Right from here, we can set
, which cancels out the
, giving
. This gives us one solution,
,
.
We can then rearrange the expression, giving us . Factoring using difference of squares, we obtain
We can then state that when is greater than
and less than
,
will be greater than
. This is obtained by setting
and then solving the inequality. We can then conclude that
.
Next, we claim that or
when
. We can prove this by first noting that since
and
are integers,
is an integer. Next, we shall assume that
. Solving this and plugging back into the original equation, we obtain
. Solving we obtain
, which is a contraction to
. Therefore,
and
or
.
Finally, we can go back to the equation
We note that since
is larger than
, in order for there to be solutions,
must be in the range
. However, this contradicts what was proven earlier, so when
and
or
, there are no solutions for
.
Now, all that remains to be checked are values of between
and
. Using brute force and checking each value individually, we can assert that there are no such solutions for
, leaving us with only
solution,
. Therefore, the answer is
Solution by treetor10145
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 39 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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