Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 6"
(Created page with "== Problem == Observe that <cmath>\begin{align*} 2^2+3^2+6^3 &= 7^2 \\ 3^2+4^2+12^3 &= 13^2 \\ 4^2+5^2+20^3 &= 21^2 \\ \end{align*}</cmath> (a) Find integers <math>x</math> a...") |
Petrichord (talk | contribs) (clarified some derivations, fixed up formatting, changed category to introductory algebra problems) |
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Observe that | Observe that | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | 2^2+3^2+6^ | + | 2^2+3^2+6^2 &= 7^2 \\ |
| − | 3^2+4^2+12^ | + | 3^2+4^2+12^2 &= 13^2 \\ |
| − | 4^2+5^2+20^ | + | 4^2+5^2+20^2 &= 21^2 \\ |
\end{align*}</cmath> | \end{align*}</cmath> | ||
| Line 15: | Line 15: | ||
(c) Prove your conjecture. | (c) Prove your conjecture. | ||
| + | == Solution == | ||
| + | (a) We can rewrite the given equation as <math>y^2-x^2 = 25+36 = 61</math>. Use difference of squares to obtain <math>(y + x)(y - x) = 61</math>. Since <math>61</math> is a prime we conclude that <math>(y + x) = 61 \text{ and } (y - x) = 1</math>, giving us <math>x = 30 \text{ and } y = 31</math>. | ||
| + | |||
| + | |||
| + | (b) It is not too hard to notice that the LHS above is <math>n^2 + (n+1)^2 + (n(n+1))^2</math> and the RHS above is <math>(n(n+1)+1)^2</math> for <math>n = 2, 3, 4 \text{ and } 5</math>. We will prove that the LHS <math>=</math> RHS for all integers (although the proof extends to real numbers) in (c). | ||
| − | == | + | (c) We expand the LHS to obtain |
| + | <cmath>\begin{align*} | ||
| + | n^2 + n^2 + 2n + 1 + n^2(n^2 + 2n + 1) &= n^4 + 2n^3 + 3n^2 + 2n + 1 \\ | ||
| + | &= (n^4 + n^3 + n^2) + (n^3 + n^2 + n) + (n^2 + n + 1) \\ | ||
| + | &= (n^2 + n + 1)^2 \\ | ||
| + | \end{align*}</cmath> | ||
| + | Thus LHS = RHS and we are done. | ||
| + | ~AK2006 | ||
== See also == | == See also == | ||
{{UNCO Math Contest box|year=1993|n=II|num-b=5|num-a=7}} | {{UNCO Math Contest box|year=1993|n=II|num-b=5|num-a=7}} | ||
| − | [[Category: | + | [[Category:Introductory Algebra Problems]] |
Latest revision as of 21:13, 19 August 2021
Problem
Observe that
(a) Find integers 
and 
so that 
(b) Conjecture a general rule that is being illustrated here.
(c) Prove your conjecture.
Solution
(a) We can rewrite the given equation as 
. Use difference of squares to obtain 
. Since 
is a prime we conclude that 
, giving us 
.
(b) It is not too hard to notice that the LHS above is 
and the RHS above is 
for 
. We will prove that the LHS 
RHS for all integers (although the proof extends to real numbers) in (c).
(c) We expand the LHS to obtain
Thus LHS = RHS and we are done.
~AK2006
See also
| 1993 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
