Irish Mathematical Olympiad

Question

Let p_1, p_2, p_3 and p_4 be four different prime numbers satisfying the equations 2p_1 + 3p_2 + 5p_3 + 7p_4 = 162, 11p_1 + 7p_2 + 5p_3 + 4p_4 = 162. Find all possible values of the product p_1p_2p_3p_4.

Accepted Answer

There are many ways to go about this. One can get upper bounds on each of

p_{1}, p_{2}, p_{3}, p_{4}

and work from there. Alternatively, you can use simple number theory ideas. As

p_{1}, p_{2}, p_{3}, p_{4}

are all different there can be at most one of them even. If they were all odd, the LHS of the first equation would be odd. Thus one of

p_{2}, p_{3}, p_{4}

is an even prime and hence equal to

2

. Now look at the second equation and we see that

p_{2}

or

p_{3}

is

2

.

Case 1.

p_{2} = 2

.

2 p_{1} + 5 p_{3} + 7 p_{4} = 152

11 p_{1} + 5 p_{3} + 4 p_{4} = 148

Subtracting gives

9 p_{1} - 3 p_{4} = - 8

which is impossible.

Case 2.

p_{3} = 2

.

2 p_{1} + 3 p_{2} + 7 p_{4} = 152

11 p_{1} + 7 p_{2} + 4 p_{4} = 152

Subtracting

9 p_{1} + 4 p_{2} - 3 p_{4} = 0

. Hence

p_{2} = 3

. That leaves two linear equations for

p_{1}

and

p_{4}

with solution

p_{1} = 5

and

p_{4} = 19

. Thus the only solution of the system of equations is

(p_{1}, p_{2}, p_{3}, p_{4}) = (5, 3, 2, 19)

. Hence the only possible value of

p_{1} p_{2} p_{3} p_{4} = 570

.