Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 6"
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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. | (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. | ||
+ | ~AK2006 | ||
== See also == | == See also == |
Revision as of 02:46, 5 October 2020
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 - = 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 + + and the RHS is 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 + + 2*n + 1 + n^2 = + 2* + 3* + 2*n + 1 on expanding and combining like terms. Now it is easy to verify that this is equal to n^2 by expanding + n + 1 squared. 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 |