Proof:  Let L(x)=mx+k.  Then L(2)=2m+k is divisible by 5 and L(5)=5m+k is divisible by 2.

L(7)=7m+k=2m+(5m+k) which is the sum of two numbers divisible by 2 so L(7) is divisible by 2.  Also, L(7)=7m+k=5m+(2m+k) which is the sum of two numbers divisible by 5 so L(7) is divisible by 5.

Then L(7) is divisible by the lcm of 2 and 5, which is 10.

Lemma QED.

Proof of the problem:

Let G(x)=P(x)-L(x).  Then G(2)=0 and G(5)=0.

Since G is a polynomial then (x-2) and (x-5) are factors of G.  So G(x) can be written as (x-2)*(x-5)*H(x).  Then G(7)=5*2*H(x)=10*H(7), thus G(7) is divisible by 10.

Since both G(7) and L(7) are divisible by 10 then G(7)+L(7)=P(7) is also divisible by 10.

Proof QED.

Note there is nothing special about 2 and 5.  Any pair of coprime nonzero integers could have been used and the same proof would apply.

"Let m and n be a pair of nonzero coprime integers.  Let p(x) be an integer polynomial such that p(m) is divisible by n and p(n) is divisible by m. Prove that p(m+n) is divisible by m*n."