Problem

Prove that every right angled triangle with sides of integral length must have the length of one of its sides a multiple of 5.

Solution

Any right angled triangle obeys Pythagoras Theorem
a2+b2=c2
Last digit of a square number depends on the last digit of the number, which was squared. 0 1 2 3 4 5 6 7 8 9
Last digit of number last digit of square
0
1
4
9
6
5
6
9
4
1
a2, b2, c2 must be end with 0, 1, 4, 5, 6, 9. We can now conside all cases.
In each case one of the sides ends on 0 or 5.
if
last digit of a2 ends with and last digit of b2 ends with then last digit of c2 must end with
1+0=1
1+1=2 Not admissable
c2 cannot end with 2
1+4=5
1+5=6
1+6=7 X
1+9=0
4+0=4
4+4=8 X
4+5=9
4+6=0
4+9=3 X
6+0=6
6+4=0
6+5=1
6+6=2 X
6+9=5