When is n² + 45 a perfect square? Can You Solve It?



When is n² + 45 equal to a perfect square, for n a positive integer? I received this MathCounts problem by email from Bill by email. Can you figure it out? Watch the video for a solution.

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28 thoughts on “When is n² + 45 a perfect square? Can You Solve It?

  1. More generally we've got n=(45-a^2)/2a, here is a generator, thanks to which we can easily find the positive integer n, inputting a=1, 3 and 5.

  2. Hey everyone, I also make math videos similar to these but they show regular solutions and solutions with tricks that are way faster, (hence my channel name)… Please stop by and consider Subscribing!! Thanks!!

  3. Spread sheet: n :: n^2 :: n^2+45 :: sqrt( n^2+45) -> look for integers. Note n=2,6,22 result in integer sqrt.

  4. 22, every square is a addition of consecutive odd numbers to the previous square, right from 1, so I found 21^2= 441+43= 484 and thus, 484+45=529= 23^2
    And i'm commenting before watching the video
    Now, after the video, I was mistaken, I found only one value of n.

  5. x is an integer as n² + 45 = (n+x)²
    (n+x)² = n² + 2xn + x²
    n² + 45 = n² +2xn + x²
    2xn + x² = 45 => x is odd
    n = (45-x²)/2x : positive integer
    x : positive if x² < 45 (x <= 5) and negative if x² > 45
    Case 1 : x = 1 => n = 22
    Case 2 : x = 3 => n = 2
    Case 3 : x = 5 => n = 2
    Case 4 : x = -7 => n = 2/7
    Case 5 : x = -9 => n = 2

    Let x = 2y + 1 where y is an integer
    n = (45 – 4y² – 4y – 1)/(4y+2)
    n = -(2y²+2y-22)/(2y+1)
    2yn + n +2y² +2y – 22 = 0
    n = -2y(n+1+y) -22

    I can't figure out anything else

    I think it's only 2 and 22

  6. idk if this is different at all but i think its easier if u say
    (n+a)^2 = n^2 + 45 for some positive integer a
    n(n+2a) = 45 and then test all values which is easier given that u only solve one equa

  7. The way I solved it was remembering that the difference between consecutive squares is an odd number…

    So, the two consecutive squares that have a difference of 45 are 484 and 529 (n=45)

    Next, if we had two squares whose roots had a difference of three, the difference would be (d-2)+d+(d+2) = 45. This totals to 3d = 45, so d = 15 and (d-2) = 13. The two consecutive squares that have a difference of 13 are 36 and 49. Making n = 6, and 36 + 45 = 81.

    Finally, for two squares whose roots had a difference of 5. We get (d-4)+(d-2)+d+(d+2)+(d+4) = 5d = 45. d = 9, so (d-4) = 5 and n = 2.

    After That, since the rule explicitly said "positive integers", I just tested n = 1 and found no equality.

    That's how I got n = 2, 6, 22

  8. My method:

    The difference between `n^2` and`(n+1)^2` is `2n+1`.

    Case 1: x=n+1. n=22 -> 2n+1 = 45.
    Case 2: x=n+3. n=6 -> `2(6)+1`+`2(7)+1`+2(8)+1` = 45.
    Case 3: x=n+5. n=2 -> `2(2)+1`+…+`2(6)+1` = 45.
    No solution exists for Case 4 or above.

    Your method is pretty cool too though.

  9. n=Sqrt(x^2-45) so I went looking for the two square numbers with a difference of 45. That is, x when squared is 45 greater than some number that can be square- rooted. So I got n=22 only and missed the other two cases 🙁 p.s. I did this in my head for about 10 seconds before watching.

  10. I did it completely differently. I noted n^2 +2n+1=(n+1)^2. So, 45=2n+1, so n=22. But then I noted that 45 could also be the sum of 3 or 5 consecutive odd numbers, so I had to look for those as well, since a square could have a non-consecutive square 45 away (but squares are always consecutive odd numbers away). For three consecutive odds, 45= 13+15+17, so let 2n+1=13, n=6. Lastly for 5 consecutive odds, 45=5+7+9+11+13, so 2n+1=5, n=2.

  11. We have n^2 + 45 = m^2, or 45 = m^2 – n^2 = (m + n)(m – n). We then find all pairs of integers whose product is 45, namely (1,45), (3,15), and (5,9). To solve for m, we add the two factors and divide by 2: (m + n + m – n)/2 = 2m/2 = m. To solve for n, we subtract the smaller from the larger and divide by 2: (m + n – (m – n))/2 = n. So for the first pair, m = (45 + 1)/2 = 23, and n = (45 – 1)/2 = 22. Likewise, m = (15 + 3)/2 = 9 and n = (15 – 3)/2 = 6. Finally, m = (9 + 5)/2 = 7 and n = (9 – 5)/2 = 2.
    For the general rule, if we want to find x such that n^2 + x = m^2, we get x = (m + n)(m – n) as above. We then find all pairs of integers whose product is x. We set the larger of each pair equal to m + n and the smaller equal to m – n. By adding the pair and dividing by two, we get m, and subtracting then dividing by two gives us n. Note that m and n must be of the same parity (both even or both odd). If not, then m+ n and m – n are odd, and dividing by two would give us fractions. Therefore x cannot be equivalent to 2 (mod 4) such as 2, 6, 10, 14, etc, because all pairs of factors of x would be of different parity.

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