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

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.

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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

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.

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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?

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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