Question

Prove there are no positive integer solutions to the equation

a^2 - b^2 = 10

Answer 1

a^2 = b^2 - 10 would be my first instinct

Answer 2

a^2=b^2+10

Answer 3

yeah, that one ;)

Answer 4

1,4,9,16,25,36 are your squares

Answer 5

a and b either be positive or negative doesn't matte

Answer 6

11, 14, 19, 26, 46, 59, ... would be the resulting sequence if we can work with it

Answer 7

Amistre64 but there is no end to the square list it can be 64,81and so on also. we gotta prove that a be an irrational

Answer 8

right, was thinking maybe a proof by induction mighta been in there someplace ...

how bout, since a and b are to be positive integers, then a,b not= 0

a^2=b^2+10

1=b^2/a^2 + 10/a^2

(b/a)^2 = -10/a^2

b/a = sqrt(-10/a^2)

b/a = sqrt(-10)/a

Answer 9

u skipped a 1 in the third step

Answer 10

its monday right? ;)

Answer 11

\[(a+b)(a-b)=10\]
\[a+b=5\]
\[a-b=2\]
\[a=\frac{7}{2}\] nope
\[a+b=10\]
\[a-b=1\]
\[a=\frac{11}{2}\] nope

Answer 12

if a in an integer and b is an integer then
\[a+b\], \[a-b\] are both integers. and there are only two ways to factor 10 as the product of two integers, namely
\[10=5\times 2\] or
\[10=10\times 1\]
since neither way gives integer a and b that is it

Answer 13

i guess i should add i am using the fact that
\[a^2-b^2=10\iff (a+b)(a-b)=10\]

Answer 14

\[\frac{a^2}{10}-\frac{b^2}{10}=1\]

is a hyperbola that fer one

Answer 15

i would say that from this: \[ a \ne b\]\[a>b\]narrows down any proving

Answer 16

and of course satellites integer factors of 10 .... i feel a bit dense today lol