- #1
recon
- 401
- 1
I've been asked to find all possible pairs (a,b) of integers such that [tex]a^3 = 5b^3[/tex].
I started by considering the cases where a and b share a common factor, so that [tex]a = kl[/tex] and [tex]b = km[/tex]. But this can be reduced to the form [tex]l^3 = 5m^3[/tex], so we can assume WLOG that a and b are coprime.
If this is the case, then [tex]5 \mid a^3[/tex], so [tex]5 \mid a[/tex]. Let [tex]a = 5x[/tex].
[tex](5x)^3 = 5b^3[/tex]
[tex]125x^3 = 5b^3[/tex]
[tex]25x^3 = b^3[/tex]
So [tex]25 \mid b^3[/tex] and [tex]5 \mid b[/tex]. Let [tex] b = 5y[/tex].
This contradicts our earlier assumption that both a and b are coprime, and hence no solution exists.
Have I made any mistakes?
I started by considering the cases where a and b share a common factor, so that [tex]a = kl[/tex] and [tex]b = km[/tex]. But this can be reduced to the form [tex]l^3 = 5m^3[/tex], so we can assume WLOG that a and b are coprime.
If this is the case, then [tex]5 \mid a^3[/tex], so [tex]5 \mid a[/tex]. Let [tex]a = 5x[/tex].
[tex](5x)^3 = 5b^3[/tex]
[tex]125x^3 = 5b^3[/tex]
[tex]25x^3 = b^3[/tex]
So [tex]25 \mid b^3[/tex] and [tex]5 \mid b[/tex]. Let [tex] b = 5y[/tex].
This contradicts our earlier assumption that both a and b are coprime, and hence no solution exists.
Have I made any mistakes?