Mathematics Asked by hardik-ddod on January 5, 2022

I am trying to determine a systematic way to find all the integer vales of $a$ such that $a^2 – 4a$ is a perfect square. If it helps, I already know that the two solutions to this equation are 0 and 4.

Furthermore, I wonder how I can prove that these are all the solutions.

We know that $a^2-4a=(a-2)^2-4$.

It is obvious that these numbers would be nonconsecutive since all consecutive perfect squares differ by odd numbers, so they are nonconsecutive. So, $(a-2)^2=4$ is the only possible outcome. This means that the only solutions are $boxed{0,4}$ as you obtained.

Hope this helped!

Answered by OlympusHero on January 5, 2022

If $mge 0$ is such that $m^2 = a^2 - 4a$ we can notice that $m^2 = a^2 - 4a< a^2 -4a + 4 =(a-2)^2 = m^2 + 4$.

So $m < a-2$. Le $(a-2)- m = k$ so that $m+k = a-2$.

so $(m+k)^2 = (a-2)^2$

$m^2 + 2mk + k^2 = a^2 -4a + 4 = m^2 + 4$

$2mk + k^2 = 4$.

but $mge 0$ and $k>0$. There's just not that many choices! $2mk ge 0$ so $k^2 = 4-2mk le 4$ so $kle 2$ so $k = 1, 2$ but $k=1$ means $2m + 1=4$ so $m=frac 32$ is not an integer. So $k =2$ and $m =0$.

That's it $m = 0$ and $m+k = a-2$ so $0+2 = a-2$ so $a=4$.

==== or ....======

If $m^2 = a^2 -4a < a^2$ then $m^2 +4 = a^2 -4a + 4 = (a-2)^2$ is also a perfect square.

Is there ever a case of two perfect squares being exact $4$ apart?

There's something that comes to mind but... I'll ignore it because clever tricks are tricks and if you don't see them right away you should be able to work it out anyway....

But if $m^2 + 4=k^2=(a-2)^2$ then ... two ideas:

- $k^2 - m^2 = (k-m)(k+m) = 4$ so either $k-m =k+m=2$ and $m=0$ and $k=2$ and $k=a-2$ so $a = 4$.

or $k-m =1$ and $k+m=4$ and $m=frac 32$ and $k=frac 52$ but that's not a perfect integer. Although we do have $(frac 32)^2 = (frac 92)^2 -4cdot frac {9}2$....

- Le $k=m+i$ and so $m^2 + 4 = k^2 =m^2 +2mi + i^2$ so $2mi+i^2 = 4$ and the only options are $i=2$ and $m=0$ so $k =2$ and $a = k+2 =4$.

.....

Clever trick that I mentioned but didn't want to use...

$n^2 = 1 + 3 + 5 + ...... +(2n-1)$ so if $k^2 - m^2 = 4$ we have a sequence of consecutive odd numbers that add to $4$. And can only be $1+3 = 4$ so $m^2 =0$ and $k^2= 1+3 = 4 = 2^2$ so $m = 0; k=2; a=k+2=4$.

Answered by fleablood on January 5, 2022

Of course you have to recognize that if $a$ is an integer, $a^2-4a+4$ is already a perfect square, that of $a-2$ and $2-a$.

But the main observation is, if you look at the sequence of squares of whole numbers (including $0$), i.e. $$0^2,1^2,2^2,3^2,4^2,5^2,6^2,7^2cdots \ 0,1,4,9,16,25,36,49,cdots$$ and look at their pairwise differences, i.e. $$1-0,4-1,9-4,16-9,25-16,36-25,49-36,cdots \ = 1,3,5,7,11,13,cdots$$ which gives precisely the odd numbers (since $(n+1)^2-n^2=2n+1$), you can deduce that to get from one square number $S_1 (a^2-4a, in this case)$ to another $S_2 (a^2-4a+4, in this case)$, you always have to * make a jump whose magnitude is the sum of a few consecutive odd numbers*.

For e.g. to move from $10^2=100$ to $16^2=256$, you have to make the jumps $16^2-15^2=31, 15^2-14^2=29, 14^2-13^2=27,13^2-12^2=25,12^2-11^2=23,11^2-10^2=21$, indeed $16^2-10^2$ can be written as a telescoping sum of the above terms.

**P.S.** - The actual fun now is sitting down and trying to figure out some obvious patterns, if any ;) for the sequence of cubes, perfect $4^{th}$ powers, perfect $5^{th}$ powers

Answered by Fawkes4494d3 on January 5, 2022

1 Asked on November 24, 2021 by zest

abstract algebra group homomorphism homological algebra quotient group tensor products

0 Asked on November 24, 2021 by silkking

1 Asked on November 24, 2021

calculus integration ordinary differential equations real analysis

1 Asked on November 24, 2021 by mathnewbie

4 Asked on November 24, 2021

eigenvalues eigenvectors linear algebra linear transformations operator algebras

1 Asked on November 21, 2021

2 Asked on November 21, 2021 by user686123

2 Asked on November 21, 2021

1 Asked on November 21, 2021 by threnody

1 Asked on November 21, 2021 by user785957

bessel functions hypergeometric function special functions summation summation method

2 Asked on November 21, 2021

2 Asked on November 21, 2021 by shaheer-ziya

0 Asked on November 21, 2021 by user297564

1 Asked on November 21, 2021 by damianodamiano

exponential distribution integration probability probability distributions

0 Asked on November 21, 2021 by physicslab1

0 Asked on November 21, 2021

0 Asked on November 21, 2021

boundary value problem laplacian linear pde mathematical physics partial differential equations

3 Asked on November 21, 2021 by shubhrajit-bhattacharya

alternative proof combinatorics number theory polynomials prime numbers

1 Asked on November 21, 2021 by isato

Get help from others!

Recent Answers

- Jon Church on Why fry rice before boiling?
- Lex on Does Google Analytics track 404 page responses as valid page views?
- Peter Machado on Why fry rice before boiling?
- haakon.io on Why fry rice before boiling?
- Joshua Engel on Why fry rice before boiling?

Recent Questions

- How Do I Get The Ifruit App Off Of Gta 5 / Grand Theft Auto 5
- Iv’e designed a space elevator using a series of lasers. do you know anybody i could submit the designs too that could manufacture the concept and put it to use
- Need help finding a book. Female OP protagonist, magic
- Why is the WWF pending games (“Your turn”) area replaced w/ a column of “Bonus & Reward”gift boxes?
- Does Google Analytics track 404 page responses as valid page views?

© 2023 AnswerBun.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP