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Title: Perfect squares Post by Altamira_64 on Apr 11th, 2012, 1:07am Consider an integer x. If we add 30, then the result is a perfect square. If we subtract 30, the result is also a perfect square. How many such integers are there?" |
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Title: Re: Perfect squares Post by pex on Apr 11th, 2012, 1:26am [hide]Two.[/hide] |
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Title: Re: Perfect squares Post by Altamira_64 on Apr 11th, 2012, 2:42am Well, I guess these must be 34 and 226. |
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Title: Re: Perfect squares Post by pex on Apr 11th, 2012, 11:12am on 04/11/12 at 02:42:54, Altamira_64 wrote:
Yes, they are. Full solution: [hideb]Let the two squares be a2 and b2, with a>b>0 without loss of generality. It is given that a2-b2 = 2*30 = 60, or equivalently (a+b)(a-b) = 60. Now we could just check all factorizations of 60 into two positive integer factors (there are only five possibilities), but let's narrow it down a little further. The difference between a+b and a-b is 2b, which is even, so the factors are both odd or both even. Two odd factors couldn't have product 60, so both factors must be even. The prime factorization is 60 = 2*2*3*5. To get two even factors, both must contain a factor 2. There are only two options for assigning the remaining primes: a) One to each factor: a+b = 2*5 = 10 and a-b = 2*3 = 6. Solving, a = 8, a2 = 64, b = 2, b2 = 4, and x = 34. b) Both to the same factor: a+b = 2*3*5 = 30 and a-b = 2. Solving, a = 16, a2 = 256, b = 14, b2 = 196, and x = 226.[/hideb] |
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Title: Re: Perfect squares Post by Altamira_64 on Apr 14th, 2012, 9:26am Great solution, Pex! Mine is different: It is based to the fact that each perfect square N^2 is the sum of the first N odd numbers (5^2 = 25 = 1+3+5+7+9). Thus the difference of 2 perfect squares should equal to the sum of consecutive odd numbers (and this should equal 60). Starting from 1, we write down the sums of the odd numbers: 1+3+5+7+9+11+13 = 49 while 1+3+5+7+9+11+13+15 = 64. Thus we cannot make 60 starting from 1. We do the same with 3: 3+5+7+9+11+13=48 while 3+5+7+9+11+13+15 = 63. Not possible. Starting from 5: 5+7+9+11+13+15=60 YESS!! Thus one perfect square is 4 and the next is 64, their difference being 60, so the first number we are looking for is 34 (34-30 = 4 and 34+30 = 64, both of them perfect squares). Similarly we find that the only other sum of consecutive odd numbers equaling 60 is 29+31. Therefore one perfect square is 1+3+5+...+27=196 (14^2) and the next is 1+3+5+...+27+29+31=256 (16^2) and the second number we are asking for is 226 (226-30 = 196, 226+30 = 256). There is no other series of successive odd numbers equaling 60, so these two are the only numbers with this property. |
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Title: Re: Perfect squares Post by SWF on Apr 16th, 2012, 3:11pm x = b2+30 = a2-30 = (a2+b2)/2 (a-b)(a+b)=60. (a-b) and (a+b) must both be even, define 2n as: 2n=a-b which means 30/n=a+b. Square these two equations, add, and divide by 4: n2 + (15/n)2 = (a2+b2)/2, but this equals the expression given above for x meaning x is the sum of the squares of any two positive integers whose product is 15. There are two ways to factor 15 into two numbers: {1,15} gives x= 12+152 = 226 {3,5} gives x= 32+52 = 34 |
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