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Question

1) Find all pairs of natural numbers that have squares that differ by 75??
2) Same as 1. But differ by 41
3) Show that there can only be one such pair for each prime number?!

Answer 1

If a^2 - b^2 = 75 then (a-b)(a+b) = 75, so by looking at all pairs of factors of 75, you can find all possibilities.
75 =
1*75 => (a-b) = 1, (a+b) = 75 => a=38, b=37
3*25 => a=14, b=11
5*15 => a=10, b=5

Now since 41 can only be factored as: 1*41, the only possibilities are:
(a-b) = 1, (a+b) = 41 =>
a=21, b=20

Similarly, any prime p can only be factored 1*p, so the only possibilities are:
(a-b) = 1, (a+b) = p =>
a=(p+1)/2, b=(p-1)/2
a and b are integers for any prime p >2

Answer 2

Let's assume our numbers are a and b, and let's say that a>b.

The difference of two squares (a^2 - b^2) can be factored into (a+b)(a-b).

So let's set THAT equal to 75.
(a-b)(a+b)=75

We are limited in that a and b are natural numbers, or "counting numbers". The addition of two counting numbers or the difference of two natural numbers will also be natural. So (a-b) and (a+b) are both natural. So let's use a factor tree on 75 to find how we can break down 75 into the multiplication of two natural numbers.

one and seventy-five
three and twenty-five
five and fifteen

If we let a=10 and b=5, then a-b will give us the five we need, and a+b will give us the fifteen we need.

Ten squared is 100, and five squared is 25, and those squares are 75 apart. So one pair of numbers is (ten and five).

I got that by starting at fifteen (the a+b) part, taking the average of that number (seven point five) and moving half the distance of the (a-b) part in both directions on the number line. So 7.5 plus 2.5 gave me ten, and 7.5 minus 2.5 gave me five. So ten and five were solutions.

For the three and twenty five solution, we start at half of twenty five which is 12.5, and move half of three in both directions on the number line, which is 1.5 places. So that gives me eleven and fourteen. When a=14 and b=11, then a-b is three, and a+b is 25.

Fourteen squared is 196 and 11 squared is 121, and they are seventy five apart.

Last case... take half of 75, and that is 37.5, move half of "one" in both directions, and that is 37 and 38. :

When you square 36, you get 1369.
Squaring 38 gives you 1444... these larger numbers are also 75 apart.

Soooo for number one, ten and five work, and 14 and 11 work and 36 and 38.

For part two, the difference is 41.

We can ONLY factor 41 one way, that is one and forty-one. Primes are like that. So by showing there is only one solution for primes, that answers part three of your homework.

(a-b)(a+b) = 41

By the method above, we have 20 and 21 as solutions and we note that the square of 21 is 441, and the square of 20 is 400, and they are indeed 41 apart.

And as primes can only be factored one way, primes defined as the difference of two squares: (a^2 - b^2) which in turn is factored (a-b)(a+b), there will only be one pair of natural numbers that have squares which differ by a given prime number.

Answer 3

(n+1)^2 - (n)^2 = 75 . . . (21, 21)
(n+1)^2 - (n)^2 = 41 . . . (37, 38)

Answer 4

with the english geniuses.

Answer 5

a

Answer 6

bob

Answer 7

MIT