MHB Is the Converse of the Given Statement True for Any Positive Integer n?

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The discussion centers on whether the difference between the smallest perfect square greater than or equal to a positive integer n (where n is prime) is itself a perfect square. Initial examples using prime numbers like 3 and 5 show that the difference can be a perfect square, suggesting that the original statement may not hold universally. Counterexamples, such as when n equals 6, indicate that the converse is also not true, as the difference is not a perfect square. Participants clarify the conditions of the problem, emphasizing that m must be greater than or equal to n. Overall, the inquiry reveals complexities in the relationship between prime numbers and perfect squares.
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if n is a positive integer greater than 2 and m the smallest integer greater than or = n, that is a perfect square.
Let a = m-n.

Show that if n is prime, then a is not a perfect square.

Also, is the converse of above true, for any integer n?

any guidance, will be much appreciated?

Thanks
 
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johnny009 said:
if n is a positive integer greater than 2 and m the smallest integer greater than or = n, that is a perfect square.
Let a = m-n.

Show that if n is prime, then a is not a perfect square.

Also, is the converse of above true, for any integer n?
any guidance, will be much appreciated?Thanks

Hey johnny009! Welcome to MHB! (Smile)Guidance: let's try a couple of examples, starting with the simplest we can think of.The smallest prime $n$ is $3$, in which case $m=2^2=4$, and $a=4-3=1$, which is a perfect square!
Ah well, maybe $a=1$ is a special case...

Let's try again, the next prime $n$ is $5$, so that $m=3^2=9$, and $a=9-5=4$, which is again a perfect square!

Erm... I think it's not true, and we have 2 counter examples to prove it.Continuing with $n=6$, we get $m=3^2=9$, and $a=9-6=3$, which is not a perfect square... and $n$ is not prime.
So we have a counter example for the converse as well.
 
I like Serena said:
Hey johnny009! Welcome to MHB! (Smile)Guidance: let's try a couple of examples, starting with the simplest we can think of.The smallest prime $n$ is $3$, in which case $m=2^2=4$, and $a=4-3=1$, which is a perfect square!
Ah well, maybe $a=1$ is a special case...

Let's try again, the next prime $n$ is $5$, so that $m=3^2=9$, and $a=9-5=4$, which is again a perfect square!

Erm... I think it's not true, and we have 2 counter examples to prove it.Continuing with $n=6$, we get $m=3^2=9$, and $a=9-6=3$, which is not a perfect square... and $n$ is not prime.
So we have a counter example for the converse as well.
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Hi There,

Thanks a lot for the reply.

But, your solutions ignores the fact, that 'm' cannot be less than 'N' ...as per the QUESTION??

So, your solution...is not really addressing the Question.

CHEERS

John.
 
johnny009 said:
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Hi There,

Thanks a lot for the reply.

But, your solutions ignores the fact, that 'm' cannot be less than 'N' ........as per the QUESTION??

So, your solution...is not really addressing the Question.

CHEERS

John.

I'm assuming you mean 'n' instead of 'N', since there is no reference to 'N'?
Erm... in each of the examples $m\ge n$ as per the question... am I missing something? (Wondering)
 
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