Can this be solved without trial and error?

[Darkmisc]

Homework Statement

One square subtracted from another square gives an area of 57. Both squares have side lengths less than 25. What is the area of the larger square?

Relevant Equations

x^2 - Y^2 = 57

Hi everyone

The following problem looks like it needs to be solved by trial and error. Is there a quicker way to do it?

I had the answers (but not solutions), so I simply worked backwards to get 11^2-8^2=57.

If I had to solve it, I would have made a list of the final digits you'd get if you squared numbers ending in 0 - 9, i.e.

0 --> 0
9 --> 1
8 --> 4
7 --> 9
6 --> 6
5 --> 5
4 --> 6
3 --> 9
2 --> 4
1 --> 1

You could also just list the squares of 1 to 25, but I'm not sure if calculators are allowed for this problem.

I'd then look at the which final digits subtracted from each other gives a final digit of 7. Here, it'd be be 1 minus 4 gives a final digit of 7.

Then, I'd use trial and error and eventually arrive at 11^2 - 8^2.

Can it be done in less steps than this?

Thanks

 

[malawi_glenn]

Factor 57 as 1*3*19
and then wlog: x≥y set x = y + z where z≥ 0 for some integer z

This will give you
x² - y² = 57
(y+z)²-y² = 58´7
y² + 2yz + z² - y² = 57
2yz + z² = 57
factorize:
z(2y + z) = 57 = 1*3*19

Now you have only a few cases to consider

case 1: z = 1 gives you 2y + 1 = 57 thus y = 28 not eligible
case 2: z = 3 gives you 2y + 3 = 19 thus y = 8 and x = 11 your solution
case 3: z = 19 gives you 2y + 19 = 3 thus y < 0 not eligible
case 4: z = 57 willl also give y< 0

When you work with integer equations, it is a good idea to factorize

 

[pasmith]

Whenever you have a difference between squares, you should start with the identity [itex]x^2 - y^2 = (x+y)(x-y)[/itex]. Then we have [tex]
(x + y)(x - y) = 57.[/tex] There are now two possibilities: [tex]
\begin{split} x + y &= \mbox{$57$ or $19$}, \\
x - y &= \mbox{$1$ or $3$}.\end{split}[/tex] This system is easliy solved to find [itex](x,y) = (29, 28)[/itex] or [itex](11,8)[/itex]. The first violates the constraint that both [itex]x[/itex] and [itex]y[/itex] are less than 25.

EDIT: An improvement is to note that the conditions of the problem require [tex]
0 < x - y < x + y < 50[/tex] so [itex]x + y = 57[/itex] is plainly impossible.

 

[malawi_glenn]

Yeah that was a much better solution than mine :) but my strategy can be used when not havin difference of squares too

 

[fresh_42]

Whenever you have a difference between squares, you should start with the identity [itex]x^2 - y^2 = (x+y)(x-y)[/itex]. Then we have [tex]
(x + y)(x - y) = 57.[/tex] There are now two possibilities: [tex]
\begin{split} x + y &= \mbox{$57$ or $19$}, \\
x - y &= \mbox{$1$ or $3$}.\end{split}[/tex] This system is easliy solved to find [itex](x,y) = (29, 28)[/itex] or [itex](11,8)[/itex]. The first violates the constraint that both [itex]x[/itex] and [itex]y[/itex] are less than 25.

 

[Mark44]

Factor 57 as 1*3*19

No need to include 1 as a factor.

This will give you
x² - y² = 57
(y+z)²-y² = 58´7

58'7 is a typo and should be 57, right?

factorize:
z(2y + z) = 57 = 1*3*19

Again, no need to include 1.

 

[malawi_glenn]

No need to include 1 as a factor.

I know but it is sometimes illustrative and helpful

58'7 is a typo and should be 57, right?

yes it is