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Which of the following are true?

1:for every positive integer m, there is a positive integer n such that m+n+1 is a perfect square.

2:for every positive integer m, there is a positive integer n such that mn+1 is a perfect cube

3:for every positive integer m, there is a positive integer n such that m+n+1 is a perfect cube

4:for every positive integer m, there is a positive integer n such that mn+1 is a perfect square

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3  
What have you tried so far? Oh, and if it's homework, tag it as such. –  ypercube Mar 14 '11 at 13:13
    
@yper: I was only allowed one (self-created) tag for my first post! Maybe a higher-up could do us the favor if/when it is confirmed as homework. –  The Chaz 2.0 Mar 14 '11 at 13:34

2 Answers 2

I'll do the first one, which should help on (3)...

Given some m $\geq 1$
Consider $(m + 1)^2 = m^2 + 2m + 1$
The left hand side $(n + m + 1)$ will equal a perfect square (namely, $m^2 + 2m + 1$) if we let $n = m^2 + m > 0$

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1  
Thanks a lot everybody but the problem was that the book gave the answer that only statement 1 and 2 are true but i was getting all of them true. i am new to such forums and shall be thankful to know how to put up questions better. –  HarshCurious Mar 15 '11 at 3:52
    
@Harsh: That's fine! Notice that I explicitly stated that our value of n was positive. When you try a similar approach to (3), can you come to such a conclusion? –  The Chaz 2.0 Mar 15 '11 at 4:01

Second is easy too. Let's assume that $mn+1$ is a perfect cube and recall that $(a + 1)^3 = a^3 + 3a^2 + 3a+ 1$. If both of them are equal we get $(mn + 1) = a^3 + 3a^2 + 3a+ 1$ which is the same as $mn = a(a^2 + 3a + 3)$. This means that for every m if you take $n = m^2 + 3m + 3$ then $mn+1$ will be a perfect cube.

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1  
$n = m^2 + 3m + 3$ you mean of course. –  ypercube Mar 14 '11 at 14:40
    
@ypercube: Thanks, fixed. –  Giorgi Mar 14 '11 at 14:47
    
And this answer will help on (4)... which should read "square" at the end?? –  The Chaz 2.0 Mar 14 '11 at 14:50

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