Add a square to $853$ to form another square

Question

Given that $853$ is a prime number, find the square number $S$ such that $S + 853$ forms another square number.

I have no idea how you would find $S$, and trial and error doesn't help.

Is there a way of finding $S$?

Hint: $a^2+n=b^2\implies n=(b-a)(b+a)$. – lulu Feb 05 '20 at 17:10 — lulu, Feb 05 '20 at 17:10
Got it. Have the answer now. Thanks – AOD Feb 05 '20 at 17:15 — AOD, Feb 05 '20 at 17:15

score 2 · Answer 1 · answered Feb 05 '20 at 17:15

Since $S$ is a square number, we can write it as $S=s^2$. Then suppose that $T$ is another square number which can be written as $T=t^2$. We then have the relation $$T=S+853\implies t^2=s^2+853\implies t^2-s^2=(t-s)(t+s)=853.$$

Since $853$ is a prime number, its only factors are $1$ and $853$, meaning that we have the linear system \begin{align} t-s&=1\\ t+s&=853 \end{align}

Can you take it from here?

Andreas Blass · Answer 2 · 2020-02-05T18:57:10.177

2

Any odd natural number $2n+1$ is the difference between two (consecutive) squares, $(n+1)^2-n^2$. Apply this with $2n+1=853$.

(If the problem had asked for the smallest $S$ of the desired sort, then you'd have to take into account that $853$ is prime, because otherwise there could be smaller solutions. But as long as nobody cares about minimizing $S$, it doesn't matter that $853$ is prime. The same method works for all odd natural numbers.)

edited Feb 05 '20 at 18:57

answered Feb 05 '20 at 18:41

Andreas Blass

71,833

$2n+1=853\ldots$ – Feb 05 '20 at 18:50
@RoddyMacPhee Thanks for the correction; I've edited it into the answer. – Andreas Blass Feb 05 '20 at 18:57

score 0 · Answer 3 · answered Feb 05 '20 at 17:15

0

Let $S=s^2$ and $S+853=T=t^2$. Then $T-S=(t+s)(t-s)=853$. But $853$ is a prime, so $t+s=853$ and $t-s=1$, yielding $s=(853-1)/2=426$ and $S=426^2=181476$.

answered Feb 05 '20 at 17:15

Parcly Taxel

103,344

score 0 · Answer 4 · answered Feb 05 '20 at 17:23

Just to algebra

Let $S = n^2$ and let $S +853 = m^2$ so

$n^2 + 853 = m^2$.

Now what?

....

Well, $m^2 - n^2 = 853$ and $(m+n)(m-n) = 853$.

Now remember... $853$ is prime and $m,n$ are natural numbers so ....

$m+n$ and $m-n$ are factors of the prime number $853$.

So $m+n = 853$ and $m-n=1$.

Two equations, two unknowns.

score 0 · Answer 5 · answered Feb 07 '20 at 16:15

Lets see :

knowing $(2x)^2=4x^2$ and $(2x+1)^2=4x^2+4x+1$ plus $853=4(213)+1$ we can conclude $S$ is of form $2x$ (if we substitute $x+1$ for $x$ and expand yhe difference is of wrong form)
knowing $(3y)^2=9y^2$ and $(3y+1)^2=9y^2+6y+1$ and finally $(3y+2)^2=9y^2+12y+4$ plus $853=3(284)+1$ means $S$ is of form $3y$ ...

Together these limit $S$ to a multiple of 6 . It also has to be greater than 29 as $29^2<853$ and lower than 428 as $428^2-427^2= 428+427=856$

Add a square to $853$ to form another square

5 Answers5