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Perfect squares that are still perfect after adding three to each digit

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Does this still have the requirement for the squares to have four digits from your previous question?
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If you add 3 to each digit of a number, you're effectively adding some number containing only 3's. For example,

    4   + 3   = 7
    16  + 33  = 29
    144 + 333 = 477

So you're looking for a perfect square, let's call it X², which becomes another perfect square, let's call it Y²

    X² + 3    =  Y²
    X² + 33   =  Y²
    X² + 333  =  Y²
    ...

We need to find some relation between the X², the Y² and the 3-number. Let's try **factorising** by first subtracting X² to the other side

         3    =  Y²-X²  =  (Y-X)·(Y+X)
         33   =  Y²-X²  =  (Y-X)·(Y+X)
         333  =  Y²-X²  =  (Y-X)·(Y+X)
         ...

Now this has become a new question:

    3    =  (Y-X)·(Y+X)
    33   =  (Y-X)·(Y+X)
    333  =  (Y-X)·(Y+X)
    ...

1. Find values for X and Y such that (Y-X)·(Y+X) equal {3,33,333,3333,...}
2. And then check if X² and Y² match the description of "adding 3 to each digit"

>!There might be a more effective way of doing this!<
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The key observation is that m^2 - n^2 = (m + n)(m - n).

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Adding 3 to each digit is the same as adding 3333, so we're looking for some n such that both n^2 and n^2 + 3333 are perfect squares between 1000 and 9999. Defining m = √(n^2 + 3333) and k = m - n (both integers), notice that m^2 - n^2 = 3333. By our key observation, we have 3333 = k(m + n). In particular, k has to be a factor of 3333, so is either 1, 3, 11, 33, 101, 303, 1111, or 3333. At this point, we could just try all of those, but I'd prefer to get it down to fewer cases first. First, note that m = n + k, so (m + n) = 2n +k, and we have 3333 = k(2n + k) = 2nk + k^(2). Rearranging that gives n = (3333/k - k)/2. Since both n^2 and m^2 have to have 4 digits, n and m need to be between 32 and 99 inclusive. Thus, k = m - n can't be any bigger than 99 - 32 = 67, so we have k = 1, 3, 11, or 33. We can still narrow that down a bit more, though: since n^2 + 3333 < 10000, we have n^2 < 6667, so n < √6667 = 82. Using our expression for n, that gives us 82 > (3333/k - k)/2, so 164 > 3333/k - k, or 0 < k^(2) + 164k - 3333. Hitting that with the quadratic formula, we get that k has to be bigger than -82 + √10057 > 18. Thus, our only possibility is k = 33. That gives n = (3333/33 - 33)/2 = (101 - 33)/2 = 68/2 = 34 and m = n + k = 34 + 33 = 67. Just to check, we have 33^2 = 1156. Adding 3333 to that gives 4489, and 67^2 = 4489.

By all of the above, this solution is unique.

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