Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given 33 natural number so that their prime divisor just with $ 7,5,2,3,11$ is formed. Prove that multiplication two number of these numbers are complete square.

Thank you.

share|cite|improve this question
Hint: factor the product. It will only have even powers... – gt6989b Dec 13 '12 at 17:41
Pigeon hole principle? You must get duplicate factors since only 5 divisors? – Cris Stringfellow Dec 13 '12 at 17:54
@gt6989b: What product? I don't understand. – TonyK Dec 13 '12 at 18:48
up vote 2 down vote accepted

Hint: Recall that the positive integer $n$ is a perfect square if and only if in the prime power factorization of $n$, each exponent is even.

Any number whose prime divisors do not include any primes other than the ones mentioned can be written as $k^2(2^a3^b5^c7^d11^e)$ where $a, b,c,d,e$ are $0$ or $1$. There are only $2^5$ sequences of $0$'s and/or $1$'s of length $5$. Now use the Pigeonhole Principle.

share|cite|improve this answer
Nicolas Can you explain $k^2(2^a3^b5^c7^d11^e)$ and what is $k^2$? – misi10 Dec 13 '12 at 18:48
Look for example at the number $2^3\cdot3^0\cdot5^7\cdot7^2\cdot11^5$. This is $(2^2\cdot5^6\cdot7^2\cdot11^4)(2^1\cdot3^0\cdot5^1\cdot7^0\cdot 11^1)$, which is $(2\cdot5^3\cdot 7\cdot 11^2)^2(2^1\cdot3^0\cdot5^17^0\cdot 11^1)$. So $k=2\cdot 5^3\cdot 7\cdot 11^2$. There may be the odd typo! – André Nicolas Dec 13 '12 at 18:56
Thanks for example. case 32+1 is one of 32 case. it is true? – misi10 Dec 13 '12 at 19:05
Because there are $33$ numbers, two of the associated strings of $0$'s and/or $1$'s will be identical. If $x$ and $y$ are numbers that give rise to identical strings of $0$'s and $1$'s, their product will only have even exponents, so will be a perfect square. Here is a ridiculously simplified example. Take $3$ powers of $5$. Then the product of two of these is a perfect square. For if $2$ or more of the exponents are even, the product of the two numbers with even exponents is a perfect square. If $2$ or more of the exponents are odd, the product of the numbers with odd exponents is a square. – André Nicolas Dec 13 '12 at 19:30
@AndréNicolas : When you explain $k^2(2^a3^b5^c7^d11^e)$ with the example it is obvious.But how did you think that the given numbers can be written in $k^2(2^a3^b5^c7^d11^e)$ this form. – sam_rox Nov 17 '14 at 14:36

Hint: you need the product to have an even number of powers of each of the primes. How many combinations of odd/even powers are possible?

share|cite|improve this answer
Can you guide me more? I don't know how to count all possibles? – misi10 Dec 13 '12 at 18:52
@misi10: it is like flipping a coin. If you had only one prime, there would be 2 possibilities: odd or even. If you had two primes, there would be 4: EE, OE, EO, OO. Can you keep going? – Ross Millikan Dec 13 '12 at 18:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.