Trick for finding the square of a number with a known square

[sepulorl000]

I was looking over the internet and I was unable to find this trick that I randomly stumbled upon when I was just randomly thinking about squares and was wondering if there was already a paper about it that I could look at? Also this is a great trick that should have been taught in high school or even middle school. I also did not know under what category this would fall into so if this post is not under this category, please let me know. Thanks!

The trick is
(a+b)(b-a)+a^2=b^2 , and this holds true as
(a+b)(b-a)+a^2=b^2
ab-a^2+b^2-ab+a^2=b^2
b^2=b^2
so if you did not have access to a calculator you could solve for 43^2 just by knowing 2^2 which would go like
(43+2)(43-2)+2^2=43^2
45*41+4=1849

 

[Steven G]

You are correct that this should have been taught in school.
Many things are taught in school but they are never used to do calculations.

Every math student should now that a^2 - b^2 = (a+b)(a-b). This is how you factor the difference of squares.
Just add b^2 to both sides and you'll get a^2 = (a+b)(a-b) + b^2 for ANY choice of b.
In calculating 53^2 you should use 3 for b, because 53-3 = 50 and multiplying a number by 50 is similar to multiplying a number by the single digit 5.
So 53^2 = (53-3)(53+3) + 3^2 = (50)(56) + 9 = 2800 + 9 = 2809.

Personally I would not compute 53^2 that way. I know that (a+b)^2 = a^2 + 2ab + b^2
I would think of (53)^2 as (50+3)^2. So a = 50 and b = 3. Then 50^2 = (50+3)^2 = 50^2 + 2(50)(3) + 3^2 =2500 + 300 + 9 = 2809. I can do this in my head. All teachers show their student's that (a+b)^2 = a^2 + 2ab + b^2 but few show their students that this works for numbers as well.

Teachers spent a whole lecture on additive inverses (3+(-3) =0, -9 + 9 = 0, -2/3 + 2/3 = 0) yet when it comes to solving (x + 3) = 0, most teachers would say to subtract 3 from both sides. Now I do not necessarily have a problem with a teacher saying to subtract 3 from both sides. I just wonder why they spend anytime on teaching additive inverses.

I have a saying that goes like this: If it works for letters, then it works for numbers. If it works for numbers, then it works for letters. If it does not work for letters, then it does not work for numbers. If it does not work for numbers, then it does not work for letters.

 

[Steven G]

(43+2)(43-2)+2^2=43^2
45*41+4=1849

Although you could, I would not use 2 for b. I would use 3 for b.
(43-3)(43+3) + 3^2 = (40)(46) + 9 = 1840 + 9 = 1849.
I have a better chance of being able to figure out (40)(46) in my head compared to (45*41). Multiplying 40 by 46 is extremely close to multiplying 4 by 46.

Here is another thing that is not always taught well in school.
All students have learned the distributed law. Well this works for numbers as well!
To multiply 4 by 46 in your head, just use the distribute law! 4(46) = 4(40 + 6) = 160 +24 = 160 + 20 + 4 = 180 + 4 = 184. So 40(46) = 1840

 

[Otis]

...if you did not have access to a calculator you could solve for 43^2 ... like ... 45*41+4=1849

If a person knows how to multiply 45x41, then why wouldn't they just multiply 43×43 to begin with?

By the way, this identity is already taught in schools:

(a+b)(a-b) = a^2 - b^2

Maybe you forgot it.

?

 

[Steven G]

If a person knows how to multiply 45x41 then why don't they just multiply 43×43 to begin?

By the way, this identity is already taught in schools:

(a+b)(a-b) = a^2 - b^2

Maybe you forgot it.

?

Otis,
Yes, this identity is taught in schools. However, it is not generally taught using numbers.
For example, 51^2-50^2 is a trivial problem using the above formula. But seriously, how many students do you know who would say 51^2-50^2 = (51 + 50)(51-50) = 101?

 

[Dr.Peterson]

I was looking over the internet and I was unable to find this trick that I randomly stumbled upon when I was just randomly thinking about squares and was wondering if there was already a paper about it that I could look at? Also this is a great trick that should have been taught in high school or even middle school. I also did not know under what category this would fall into so if this post is not under this category, please let me know. Thanks!

The trick is
(a+b)(b-a)+a^2=b^2 , and this holds true as
(a+b)(b-a)+a^2=b^2
ab-a^2+b^2-ab+a^2=b^2
b^2=b^2
so if you did not have access to a calculator you could solve for 43^2 just by knowing 2^2 which would go like
(43+2)(43-2)+2^2=43^2
45*41+4=1849

There are, of course, many "tricks" you could use; the hard part is to pick one that will actually save time. Often, we really use tricks mostly because they are impressive.

Since I don't happen to know 45*41 offhand, I might do something like what Jomo suggested for 53, and use 7:

43^2 - 7^2 = (43+7)(43-7) = 50*36, so 43^2 = 50*36 + 49 = 100*18 + 49 = 1849............................................ corrected​

which is very pretty! (Note that I didn't memorize your idea as a formula, but just used what I have already memorized, that a^2 - b^2 = (a+b)(a-b).)

The good thing you are illustrating is the value of "randomly thinking" about math, rather than just sticking with what you've been taught. That is really what schools should teach -- don't just learn, discover! It's what worked for me.

 

[Steven G]

A mathematician saying don't just learn, discover! It's what worked for me is just so classic. I think that the probability of finding a mathematician who disagrees with that statement will be a true 0 (no limit is needed).