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Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 6"

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== Solution ==
 
== Solution ==
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(a) We can rewrite the given equation as <math>x^2</math> - <math>y^2</math> = 71. Use difference of squares to obtain (x + y) *  (x - y) = 71. Since 71 is a prime we conclude that (x + y) = 71 and (x - y) = 1. This gives us x = 36 and y = 35. We can verify that this is correct on substituting these values in the original equation.  
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(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>.
  
  
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(b) It is not too hard to notice that the LHS is <math>n^2</math> + <math>(n+1)^2</math> + <math>(n*(n+1))^2</math> and the RHS is <math>(n*(n+1)+1)^2</math> for n = 2, 3, 4 and 5. We will prove that LHS = RHS for all n in integers in (c).
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(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).
  
  
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(c) We expand the LHS to obtain <math>n^2</math> + <math>n^2</math> + 2*n + 1 + <math>(</math>n^2<math>+n)^2</math> = <math>n^4</math> + 2*<math>n^3</math> + 3*<math>n^2</math> + 2*n + 1 on expanding and combining like terms. Now it is easy to verify that this is equal to <math>(</math>n^2<math>+n+1)^2</math> by expanding <math>n^2</math> + n + 1 squared. Thus LHS = RHS and we are done.
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(c) We expand the LHS to obtain  
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<cmath>\begin{align*}
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n^2 + n^2 + 2n + 1 + n^2(n^2 + 2n + 1) &= n^4 + 2n^3 + 3n^2 + 2n + 1 \\
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&= (n^4 + n^3 + n^2) + (n^3 + n^2 + n) + (n^2 + n + 1) \\
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&= (n^2 + n + 1)^2 \\
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\end{align*}</cmath>  
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Thus LHS = RHS and we are done.
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~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}}
  
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[[Category:Intermediate Number Theory Problems]]
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[[Category:Introductory Algebra Problems]]

Latest revision as of 21:13, 19 August 2021

Problem

Observe that \begin{align*} 2^2+3^2+6^2 &= 7^2 \\ 3^2+4^2+12^2 &= 13^2 \\ 4^2+5^2+20^2 &= 21^2 \\ \end{align*}

(a) Find integers $x$ and $y$ so that $5^2+6^2+x^2=y^2.$

(b) Conjecture a general rule that is being illustrated here.

(c) Prove your conjecture.

Solution

(a) We can rewrite the given equation as $y^2-x^2 = 25+36 = 61$. Use difference of squares to obtain $(y + x)(y - x) = 61$. Since $61$ is a prime we conclude that $(y + x) = 61 \text{ and } (y - x) = 1$, giving us $x = 30 \text{ and } y = 31$.


(b) It is not too hard to notice that the LHS above is $n^2 + (n+1)^2 + (n(n+1))^2$ and the RHS above is $(n(n+1)+1)^2$ for $n = 2, 3, 4 \text{ and } 5$. 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 \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*} 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
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