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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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(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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(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.
  
 
== See also ==
 
== See also ==

Revision as of 03:45, 5 October 2020

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 $x^2$ - $y^2$ = 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.


(b) It is not too hard to notice that the LHS is $n^2$ + $(n+1)^2$ + $(n*(n+1))^2$ and the RHS is $(n*(n+1)+1)^2$ for n = 2, 3, 4 and 5. We will prove that LHS = RHS for all n in integers in (c).


(c) We expand the LHS to obtain $n^2$ + $n^2$ + 2*n + 1 + $($n^2$+n)^2$ = $n^4$ + 2*$n^3$ + 3*$n^2$ + 2*n + 1 on expanding and combining like terms. Now it is easy to verify that this is equal to $($n^2$+n+1)^2$ by expanding $n^2$ + n + 1 squared. Thus LHS = RHS and we are done.

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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