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Difference between revisions of "1977 Canadian MO Problems/Problem 1"
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Thus, <math>4a^2 + 4a = b^2 + b</math>, or <math>(2a+1)^2 = b^2 + b + 1</math>. But <math>b^2 < b^2 + b + 1 < (b+1)^2</math>, so <math>b^2 + b + 1</math> cannot be a perfect square, contradiction. The conclusion follows.
 
Thus, <math>4a^2 + 4a = b^2 + b</math>, or <math>(2a+1)^2 = b^2 + b + 1</math>. But <math>b^2 < b^2 + b + 1 < (b+1)^2</math>, so <math>b^2 + b + 1</math> cannot be a perfect square, contradiction. The conclusion follows.
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==Alternate Solutions?==
 
{{alternate solutions}}
 
{{alternate solutions}}
  

Revision as of 22:32, 21 September 2014

Problem

If $f(x)=x^2+x,$ prove that the equation $4f(a)=f(b)$ has no solutions in positive integers $a$ and $b.$

Solution

Directly plugging $a$ and $b$ into the function, $4a^2+4a=b^2+b.$ We now have a quadratic in $a.$

Applying the quadratic formula, $a=\frac{-1\pm \sqrt{b^2+b+1}}{2}.$

In order for both $a$ and $b$ to be integers, the discriminant must be a perfect square. However, since $b^2< b^2+b+1 <(b+1)^2,$ the quantity $b^2+b+1$ cannot be a perfect square when $b$ is an integer. Hence, when $b$ is a positive integer, $a$ cannot be.

Solution

Suppose there exist positive integral $a$ and $b$ such that $4f(a) = f(b)$.

Thus, $4a^2 + 4a = b^2 + b$, or $(2a+1)^2 = b^2 + b + 1$. But $b^2 < b^2 + b + 1 < (b+1)^2$, so $b^2 + b + 1$ cannot be a perfect square, contradiction. The conclusion follows.

Alternate Solutions?

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

1977 Canadian MO (Problems)
Preceded by
First question 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 2
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